This chapter is devoted to the investigation of infinite sequences. We will be particularly concerned how to prove the convergence (or divergence) of a sequence and to calculate its limit.

A **numerical sequence** is an ordered (or, in other words, enumerated) collection of numbers. There is no restriction for these numbers: they can be different or appear multiple times. An **infinite sequence** has infinitely many elements, denoted usually by *a*
_{n}, starting from the first one, i.e., *a*
_{1}.

The sequence is **increasing** if Similarly for a **decreasing** sequence one has If only inequalities *a*
_{n+1} ≥ *a*
_{n} or *a*
_{n+1} ≤ *a*
_{n} hold, the sequences are called **nondecreasing** or **nonincreasing** respectively. Sequences with either of these properties are called **monotonic**.

$\displaystyle \begin{aligned} \forall_{n\in \mathbb{N}}\;\; a_{n+1}>a_n. {} \end{aligned} $

(5.0.1)

$\displaystyle \begin{aligned} \forall_{n\in \mathbb{N}}\;\; a_{n+1}<a_n. {} \end{aligned} $

(5.0.2)

The sequence is **bounded above** if and **bounded below** if

$\displaystyle \begin{aligned} \exists_{M\in\mathbb{R}}\forall_{n\in \mathbb{N}}\;\; a_{n}\leq M, {} \end{aligned} $

(5.0.3)

$\displaystyle \begin{aligned} \exists_{M\in\mathbb{R}}\forall_{n\in \mathbb{N}}\;\; a_{n}\geq M. {} \end{aligned} $

(5.0.4)

A **subsequence** of a certain sequence is created by omitting some of its terms without changing the ordering of the remaining ones. For instance *b*
_{n} = *a*
_{2n} denotes the subsequence created by deleting all odd-numbered terms of the sequence *a*
_{n}.

The sequence can be specified by an explicit formula for *a*
_{n}, which can be treated as a function *a*(*n*), whose domain is the set of natural numbers $\mathbb N$, or by **recursion**. In this latter case, one has to fix the value of the first element *a*
_{1} and provide the dependence *a*
_{n+1} = *f*(*a*
_{n}). The more complicated dependence, for instance *a*
_{n+2} = *f*(*a*
_{n+1}, *a*
_{n}), requires fixing two first elements, *a*
_{1} and *a*
_{2}, etc.

The principal notion for the infinite sequence is its limit. A sequence is called **convergent** to a certain number *g* if If such number *g* exists it is called the **limit** of the sequence, which is written as If it does not, the sequence is **divergent**. A sequence that is bounded and monotonic is convergent. There are several tests for the investigation of the convergence of sequences which are discussed in detail when solving particular problems.

$\displaystyle \begin{aligned} \forall_{\epsilon>0}\exists_{N\in \mathbb{N}}\forall_{n>N}\;\; |a_n-g|<\epsilon. {} \end{aligned} $

(5.0.5)

$\displaystyle \begin{aligned} \lim_{n\rightarrow\infty} a_n=g. {} \end{aligned} $

(5.0.6)

For convergent sequences, the following equations hold: and inserting into the second equation a constant sequence for *b*
_{n}, i.e., *b*
_{n} = *c*, one also gets the relation lim_{n→∞}*c* *a*
_{n} = *c*lim_{n→∞}*a*
_{n}.

lim

_{n→∞}*a*_{n}±lim_{n→∞}*b*_{n}=lim_{n→∞}(*a*_{n}±*b*_{n}),(lim

_{n→∞}*a*_{n}) ⋅ (lim_{n→∞}*b*_{n}) =lim_{n→∞}(*a*_{n}⋅*b*_{n}),(lim

_{n→∞}*a*_{n})∕(lim_{n→∞}*b*_{n}) =lim_{n→∞}(*a*_{n}∕*b*_{n}), if lim_{n→∞}*b*_{n}≠ 0,

It can also be proved that a sequence in the special form where and *c*
_{n} satisfies the condition has the following limit:

$\displaystyle \begin{aligned} a_n=(1+b_n)^{c_n} , {}\end{aligned} $

(5.0.7)

$\displaystyle \begin{aligned}\lim_{n\rightarrow\infty} b_n=0\end{aligned} $

$\displaystyle \begin{aligned} \lim_{n\rightarrow\infty} b_nc_n=g\neq\pm\infty , {}\end{aligned} $

(5.0.8)

$\displaystyle \begin{aligned} \lim_{n\rightarrow \infty} a_n=e^g. {} \end{aligned} $

(5.0.9)

A number *a* is called the **cluster point** of a sequence *a*
_{n} if there exists a certain subsequence convergent to *a*. The **extremal limits** of a sequence are defined as follows. i.e., they are the lower and upper bounds of the set of all cluster points. For convergent sequences, both extremal limits are equal to each other and to the limit of the sequence.

- The
**lower extremal limit**:$\displaystyle \begin{aligned} \liminf _{n\rightarrow \infty }a_{n}:=\lim _{n\rightarrow \infty}(\inf_{k>n}a_{k}), {} \end{aligned} $(5.0.10) - the
**upper extremal limit**:$\displaystyle \begin{aligned} \limsup _{n\rightarrow \infty }a_{n}:=\lim _{n\rightarrow \infty}(\sup_{k>n}a_{k}), {} \end{aligned} $(5.0.11)

The convergence of the sequence will be proved and its limit will be found.

$\displaystyle \begin{aligned} a_n=\sqrt[4]{n^4+2n^3}-\sqrt[4]{n^4+n^3} {} \end{aligned} $

(5.1.1)

The sequence *a*
_{n} is typical for problems where one has to find the limit of the kind *∞*−*∞*. The procedure in all these cases is very similar. It comprises the following steps:

- 1.
First one should have to look at both the diverging terms in

*a*_{n}to assess whether their degrees of divergence are identical. A chance (but not certainty) for the finite limit exists only when leading terms behave identically for*n*→*∞*, since this allows for the cancellation of troublesome expressions. - 2.
If these terms actually behave in the same way, one should, with the use of appropriate transformations, bring about their real deletion. Then, most often, the limit will no longer be of the type

*∞*–*∞*because—one can say—the infinities “subtracted” from each other. - 3.
At the end, one finds the limit of simplified expression.

A similar procedure, still involving the separation of leading terms, will be met again in the calculation of limits in Problem 4 of the current section and Problem 1 in Sect. 5.5, and now we shall pursue it for *a*
_{n} defined by (5.1.1).

Let us look at the first term. Under the fourth root the polynomial *n*
^{4} + 2*n*
^{3} is found. For very large *n* the leading term is *n*
^{4}, so one can expect that $\sqrt [4]{n^4+n^3}\simeq \sqrt [4]{n^4}=n$. The behavior of the second term is similar. One can, therefore, hope for cancellation of both divergent expressions and for the finite limit of *a*
_{n}.

One should be aware that the above reasoning is not strict and does not prejudge the existence of the limit. Even if leading terms do cancel, this limit still may not exist. Equally true, it may exist and be finite or even be equal to zero. Examples of such different behaviors for apparently similar sequences will be given below. Despite the lack of full strictness in reasoning carried out above, it plays an important role: it indicates to us the possible solution to the problem. It is, as already mentioned, the precise cancellation of unwanted expressions.

How then can we accurately reduce the diverging terms? To do this, of course, we have to get rid of the symbols of roots. This can be done if one recalls the familiar short multiplication formula (*a* − *b*)(*a* + *b*) = *a*
^{2} − *b*
^{2}. Let our *a* and *b* be, respectively, the first and the second terms in the formula for *a*
_{n}. Then one can write For now, one has failed to reduce the unwanted terms, but still one has succeeded: instead of roots of the fourth degree in the numerator the square roots have occurred. This means that if one repeats this procedure once again both roots in the numerator will disappear and the expected cancellation will take place. One should also note here that roots in the denominator do not pose any problem, because they *add* to each other, and not *subtract*!

$\displaystyle \begin{aligned} \begin{array}{rcl} a_n &\displaystyle =&\displaystyle \sqrt[4]{n^4+2n^3}-\sqrt[4]{n^4+n^3}\\ &\displaystyle =&\displaystyle (\sqrt[4]{n^4+2n^3}-\sqrt[4]{n^4+n^3})\;\frac{\sqrt[4]{n^4+2n^3}+\sqrt[4]{n^4+n^3}}{\sqrt[4]{n^4+2n^3}+\sqrt[4]{n^4+n^3}}\\ &\displaystyle =&\displaystyle \frac{\sqrt{n^4+2n^3}-\sqrt{n^4+n^3}}{\sqrt[4]{n^4+2n^3}+\sqrt[4]{n^4+n^3}}.{} \end{array} \end{aligned} $

(5.1.2)

Let us then write

$\displaystyle \begin{aligned} \begin{array}{rcl} a_n&\displaystyle =&\displaystyle \frac{\sqrt{n^4+2n^3}-\sqrt{n^4+n^3}}{\sqrt[4]{n^4+2n^3}+\sqrt[4]{n^4+n^3}}\cdot\frac{\sqrt{n^4+2n^3}+\sqrt{n^4+n^3}}{\sqrt{n^4+2n^3}+\sqrt{n^4+n^3}}\\ &\displaystyle =&\displaystyle \frac{(n^4+2n^3)-(n^4+n^3)}{(\sqrt[4]{n^4+2n^3}+\sqrt[4]{n^4+n^3})(\sqrt{n^4+2n^3}+\sqrt{n^4+n^3})}{}\\ &\displaystyle =&\displaystyle \frac{n^3}{(\sqrt[4]{n^4+2n^3}+\sqrt[4]{n^4+n^3})(\sqrt{n^4+2n^3}+\sqrt{n^4+n^3})}. \end{array} \end{aligned} $

(5.1.3)

This quotient expression is much easier to examine. One needs only to extract the highest power of *n* in the numerator and in the denominator and one can calculate the required limit: Its value has been established, having regard to the following items:

$\displaystyle \begin{aligned} \lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}\frac{n^3}{n^3}\cdot\frac{1}{(\sqrt[4]{1+2/n}+\sqrt[4]{1+1/n})(\sqrt{1+2/n}+\sqrt{1+1/n})} = \frac{1}{4}. {} \end{aligned} $

(5.1.4)

$\displaystyle \lim _{n\rightarrow \infty }\frac {n^m}{n^m}=1$.

$\displaystyle \lim _{n\rightarrow \infty }\sqrt [k]{1+a/n}=\sqrt [k]{\lim _{n\rightarrow \infty }(1+a/n)}=\sqrt [k]{1}=1$, where the first equality follows from the continuity of the “root” function, thanks to which the limit can be moved into the argument.

- The following operations can be carried out on the limits of sequences:if all these limits exist and, in the latter case, the limit of$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{n\rightarrow\infty}(b_n+c_n)&\displaystyle =&\displaystyle \lim_{n\rightarrow\infty}b_n+\lim_{n\rightarrow\infty}c_n ,\\ \lim_{n\rightarrow\infty}(b_n\cdot c_n)&\displaystyle =&\displaystyle \lim_{n\rightarrow\infty}b_n\cdot\lim_{n\rightarrow\infty}c_n ,\\ \lim_{n\rightarrow\infty}\frac{b_n}{c_n}&\displaystyle =&\displaystyle \frac{\displaystyle\lim_{n\rightarrow\infty}b_n}{\displaystyle\lim_{n\rightarrow\infty}c_n}\;, \end{array} \end{aligned} $
*c*_{n}(and, therefore, almost all its terms) is not equal to zero.

At the end, it is worth emphasizing that in this type of example, it is nonleading terms (that is, those that remain after cancellation of diverging expressions) that are essential for the existence of the limit and its value. Let us, for example, make apparently unimportant changes in the definition of sequence and write: It would seem that additional $\sqrt {n}$ is of no concern to the limit because the behavior of the sequence is determined by *n*
^{4}. Remember, however, that it is precisely this main term which finally disappears. Proceeding in the same way as in (5.1.2) and (5.1.3), we get:

$\displaystyle \begin{aligned} a_n=\sqrt[4]{n^4+2\sqrt{n}\cdot n^3}-\sqrt[4]{n^4+\sqrt{n}\cdot n^3}. {} \end{aligned} $

(5.1.5)

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle a_n= \frac{\sqrt{n}\cdot n^3}{n^3}{}\\ &\displaystyle &\displaystyle \cdot \frac{1}{(\sqrt[4]{1+2/\sqrt{n}}+\sqrt[4]{1+1/\sqrt{n}})(\sqrt{1+2/\sqrt{n}}+\sqrt{1+1/ \sqrt{n}})}\underset{n\rightarrow \infty}{\longrightarrow}\infty. \end{array} \end{aligned} $

(5.1.6)

The other seemingly unimportant modification of the formula for *a*
_{n}: leads, in turn, to the limit equal to zero: One has then to be very careful while formulating conclusions concerning the existence and values of limits only on the basis of leading terms.

$\displaystyle \begin{aligned} a_n=\sqrt[4]{n^4+2\cdot n^2}-\sqrt[4]{n^4+ n^2} , {} \end{aligned} $

(5.1.7)

$\displaystyle \begin{aligned} a_n=\frac{n^2}{n^3}\cdot\frac{1}{(\sqrt[4]{1+2/n^2}+\sqrt[4]{1+1/n^2})(\sqrt{1+2/n^2}+\sqrt{1+1/n^2})}\underset{n\rightarrow \infty}{\longrightarrow} 0. {} \end{aligned} $

(5.1.8)

The convergence of the sequence will be proved and its limit will be found.

$\displaystyle \begin{aligned} a_n=\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\ldots+\frac{1}{\sqrt{n^2+n}} {} \end{aligned} $

(5.1.9)

As usual, we are going to start solving the problem by carefully looking at the formula for *a*
_{n}. It is easy to realize that when *n* →*∞* each of the terms separately tends to zero. However, it would be a serious mistake to conclude that *a*
_{n} goes to zero as well. Why? Well, because with the decrease of individual terms their total number increases. The conclusion that $a_n\underset {n\rightarrow \infty }{\longrightarrow }0 $ might actually be drawn if the quantity of ingredients remained bounded. It would be true, for example, if we considered a sequence whose *n*-th term would have the form for $k\in \mathbb {N}$ fixed. On the contrary, for *a*
_{n} defined by (5.1.9), the value of the limit is the result of the interplay between the number of terms and the rate of their convergence to zero.

$\displaystyle \begin{aligned} \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\ldots+\frac{1}{\sqrt{n^2+k}} , {} \end{aligned} $

(5.1.10)

One of the standard ways of dealing with limits of that kind is to use the so-called “squeeze theorem.” It says that if two sequences *b*
_{n} and *c*
_{n} are found with a common limit *g*, satisfying the inequality *b*
_{n} ≤ *a*
_{n} ≤ *c*
_{n} for almost all *n*, then lim_{n→∞}*a*
_{n} = *g*.

Thus, it remains now to choose sequences *b*
_{n} and *c*
_{n}. How can one guess them? Well, they must converge to the common limit, which will also be the limit of the tested sequence *a*
_{n}. Therefore, we should have at least a certain idea of what the value of that limit would be. Finding the answer to this question is our first task.

When looking at *a*
_{n}, one sees that for large *n* each of the terms behaves as $1/\sqrt {n^2} = 1/n$, and their total number is equal to *n*. We suspect, therefore, that lim_{n→∞}*a*
_{n} =lim_{n→∞}*n*∕*n* = 1 = *g*. It remains to strictly demonstrate this.

Now one has to look for *b*
_{n} and *c*
_{n}. We already know that they should converge to *g* = 1. In the formula (5.1.9), all terms are in descending order. This suggests a certain idea: let us replace all *n* terms in the sum with the largest one, that is the first one, and a sequence obtained in this way will be called *c*
_{n}: Obviously $\forall _{n\in \mathbb {N}}, \; \; a_n\leq c_n$ and, notably Similarly, one can find sequence *b*
_{n}, this time by *n*-fold duplication of the smallest term: Naturally one has $\forall _{n\in \mathbb {N}},\;\; b_n\leq a_n$ and In this way, it is clear that the assumptions of the squeeze theorem are satisfied. The conclusion is then the lim_{n→∞}*a*
_{n} = 1.

$\displaystyle \begin{aligned} c_n=\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+1}}+\ldots+\frac{1}{\sqrt{n^2+1}}=n\;\frac{1}{\sqrt{n^2+1}}. {} \end{aligned} $

(5.1.11)

$\displaystyle \begin{aligned} \lim_{n\rightarrow\infty}c_n=\lim_{n\rightarrow\infty}\frac{n}{n}\cdot\frac{1}{\sqrt{1+1/n^2}}=1. {} \end{aligned} $

(5.1.12)

$\displaystyle \begin{aligned} b_n=\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+n}}+\ldots+\frac{1}{\sqrt{n^2+n}}=n\;\frac{1}{\sqrt{n^2+n}}. {} \end{aligned} $

(5.1.13)

$\displaystyle \begin{aligned} \lim_{n\rightarrow\infty}b_n=\lim_{n\rightarrow\infty}\frac{n}{n}\cdot\frac{1}{\sqrt{1+1/n}}=1. {} \end{aligned} $

(5.1.14)

It is worth noting that the key to the solution was the fact that all of the components of (5.1.9) behave for large *n* identically (as 1∕*n* ). If certain seemingly minor changes in the definition of *a*
_{n} are made and one writes then the difficulties in finding the limit will be much more serious. When *n* →*∞*, the first term and the last term tend to zero as 1∕*n*, and as $ 1/(n\sqrt {2})$ respectively. Therefore, one is not able to predict the limit. A purely “mechanical” application of previous definitions gives Hence the sequences do not have a common limit, and the squeeze theorem cannot be applied.

$\displaystyle \begin{aligned} a_n=\frac{1}{\sqrt{n^2+1^2}}+\frac{1}{\sqrt{n^2+2^2}}+\ldots+\frac{1}{\sqrt{n^2+n^2}} , {} \end{aligned} $

(5.1.15)

$\displaystyle \begin{aligned} b_n = n\;\frac{1}{\sqrt{n^2+n^2}}\underset{n\rightarrow \infty}{\longrightarrow} \frac{1}{\sqrt{2}} ,\;\;\;\;\mathrm{and}\;\;\;\; c_n = n\;\frac{1}{\sqrt{n^2+1^2}}\underset{n\rightarrow \infty}{\longrightarrow} 1. {} \end{aligned} $

(5.1.16)

The convergence of the sequence where $k\in \mathbb {N}$, and *λ*
_{i} > 0, *i* = 1, 2, …, *k*, will be proved and its limit will be found.

$\displaystyle \begin{aligned} a_n=\sqrt[n]{\lambda_1^n+\lambda_2^n+\ldots+\lambda_k^n} , {}\end{aligned} $

(5.1.17)

Finding the above limit is a typical exercise for the application of the squeeze theorem. Just as in the previous example, we have to consider above all whether it is possible to formulate any predictions as to the value of the limit. This is necessary for choosing the two auxiliary sequences, which—as we know—both must converge to it.

For a given *n* in the sum under the root, there is a fixed number of components equal to *k*. One can assume then that among them there is a largest and a smallest. Possibly there may even be a few if *λ*
_{i}’s are not all different for various *i*, but this has no effect on our reasoning. Let us denote The exponentials of the type *λ*
^{n} quickly grow (for *λ* > 1) or decrease (for *λ* < 1 ); therefore, the behavior of the whole expression for large *n* shall be determined by the largest number of all *λ*
_{i}’s. Therefore, we suspect that Assuming that the value of the limit has been guessed correctly, one must now indicate two sequences *b*
_{n} and *c*
_{n}, still convergent to *λ*
_{max} and satisfying—for almost all *n*—the inequalities *b*
_{n} ≤ *a*
_{n} ≤ *c*
_{n}.

$\displaystyle \begin{aligned} \lambda_{\mathrm{max}}=\mathrm{max}\{\lambda_1, \lambda_2, \ldots,\lambda_k\} ,\;\;\;\;\;\; \lambda_{\mathrm{min}}=\mathrm{min}\{\lambda_1, \lambda_2, \ldots,\lambda_k\}. {} \end{aligned} $

(5.1.18)

$\displaystyle \begin{aligned} a_n\simeq \sqrt[n]{\lambda^n_{\mathrm{max}}}=\lambda_{\mathrm{max}}\underset{n\rightarrow \infty}{\longrightarrow}\lambda_{\mathrm{max}}. {}\end{aligned} $

(5.1.19)

The choice of *b*
_{n} is clear. Since it is anticipated that the limit of *a*
_{n} is determined by the greatest number of *λ*
_{i}’s only, then the same value of the limit should be obtained if one disregards all other terms under the root. We are going to propose, therefore, Since all omitted terms were positive, obviously the inequality *b*
_{n} < *a*
_{n} holds. In turn, in order to select *c*
_{n}, we remember that for any fixed positive *α* one has $\displaystyle \lim _{n\rightarrow \infty }\sqrt [n]{\alpha }=1$. This suggests that one might put all *λ*
_{i}’s equal to *λ*
_{max}: Of course *a*
_{n} < *c*
_{n} and the squeeze theorem gives the expected result: lim_{n→n}*a*
_{n} = *λ*
_{max}.

$\displaystyle \begin{aligned} b_n=\sqrt[n]{0+\ldots+0+\lambda_{\mathrm{max}}^n+0+\ldots+0}= \lambda_{\mathrm{max}}. {} \end{aligned} $

(5.1.20)

$\displaystyle \begin{aligned} c_n=\sqrt[n]{\lambda_{\mathrm{max}}^n+\ldots+\lambda_{\mathrm{max}}^n+\ldots+\lambda_{\mathrm{max}}^n}= \sqrt[n]{k\lambda_{\mathrm{max}}^n}=\sqrt[n]{k}\,\lambda_{\mathrm{max}}\underset{n\rightarrow \infty}{\longrightarrow}\lambda_{\mathrm{max}}. {} \end{aligned} $

(5.1.21)

It is worth noticing that estimates similar to those of the previous example (i.e., the sum of *k* smallest terms and the sum of *k* largest ones) would fail. For, leaving the string *c*
_{n} without changes, and choosing *b*
_{n} as one gets If not all *λ*
_{i} are equal, then *b*
_{n} and *c*
_{n} do not converge to the common limit. When solving these examples, one should realize that not only the selection of the appropriate criteria (in this case, the squeeze theorem), but also the way of further proceeding is dictated by the special form of the sequence under study.

$\displaystyle \begin{aligned} b_n=\sqrt[n]{\lambda_{\mathrm{min}}^n+\ldots+\lambda_{\mathrm{min}}^n+\ldots+\lambda_{\mathrm{min}}^n} , {} \end{aligned} $

(5.1.22)

$\displaystyle \begin{aligned} b_n= \sqrt[n]{k\lambda_{\mathrm{min}}^n}=\sqrt[n]{k}\lambda_{\mathrm{min}}\underset{n\rightarrow \infty}{\longrightarrow}\lambda_{\mathrm{min}}. {}\end{aligned} $

(5.1.23)

The limit of the sequence where *a* > 0, will be found.

$\displaystyle \begin{aligned} a_n=\sin\left(\pi\sqrt{a^2+n^2}\right) , {}\end{aligned} $

(5.1.24)

In the first exercise of this section, we already encountered the situation where leading terms diverging to infinity canceled, and a finite “remainder” determined the limit. A similar case is dealt with in the present example. Someone might ask where the place for cancellation of infinities is if in our formula for *a*
_{n}’s, there appears no subtraction of any divergent expressions at all. Well, this cancellation results from the periodicity of the sine function. It is true that the argument of this function goes to infinity, but one can always reduce it to the interval [0, 2*π*[ by subtracting from it the appropriate multiple of the period equal to 2*π*. For the sine function, then, it is not the value of the argument itself that is important, but rather the remainder after this reduction. This is reflected in our further steps.

If one looks at *a*
_{n}, one can immediately see that, for *n* large enough, the argument of the sine function behaves as $\pi \sqrt [2]{n^2}=n\pi $. That is just the value that should be removed from the argument. This can be done by writing Using now the formula for the sine of the sum of two angles, one gets where the following properties of trigonometric functions have been used: $\sin n\pi =0$ and $\cos n\pi = (-1)^n$. If one now looks at the formula for the general term *a*
_{n} which is obtained, one sees that it is a product of two factors: (−1)^{n}, which is bounded, and $\sin \left (\pi \sqrt {a^2+n^2} -n\pi \right )$, the limit of which still has to be found.

$\displaystyle \begin{aligned} a_n=\sin\left(\pi\sqrt{a^2+n^2}\right)=\sin\left(\pi\sqrt{a^2+n^2}-n\pi+n\pi\right). {}\end{aligned} $

(5.1.25)

$\displaystyle \begin{aligned} \sin{}(\alpha+\beta)=\sin\alpha\cos\beta+ \sin\beta\cos\alpha ,\end{aligned}$

$\displaystyle \begin{aligned} \begin{array}{rcl} a_n&\displaystyle =&\displaystyle \sin\left(\pi\sqrt{a^2+n^2}-n\pi\right)\cos n\pi+\cos\left(\pi\sqrt{a^2+n^2}-n\pi\right)\sin n\pi\\ &\displaystyle =&\displaystyle (-1)^n\sin\left(\pi\sqrt{a^2+n^2}-n\pi\right) , {} \end{array} \end{aligned} $

(5.1.26)

The sine function is continuous, so we can consider the limit of the argument. How to proceed with the expression $ \sqrt {a^2+n^2} -n$ is already known from the first example in this section:

$\displaystyle \begin{aligned} \begin{array}{rcl} \sqrt{a^2+n^2}-n&\displaystyle =&\displaystyle (\sqrt{a^2+n^2}-n)\;\frac{\sqrt{a^2+n^2}+n}{\sqrt{a^2+n^2}+n}=\frac{a^2+n^2-n^2}{\sqrt{a^2+n^2}+n}\\ &\displaystyle =&\displaystyle \frac{a^2}{\sqrt{a^2+n^2}+n}=\frac{a^2}{n}\cdot\frac{1}{\sqrt{1+a^2/n^2}+1}\underset{n\rightarrow \infty}{\longrightarrow}0.\\ {} \end{array} \end{aligned} $

(5.1.27)

Since $\sin 0 = 0$ in the equation (5.1.26), one has the product of a bound expression, and that convergent to zero. The product of this type is convergent to zero as well, which can be easily justified. Adopting such notation that *b*
_{n} means a bounded expression (by a number *M* > 0) and *c*
_{n} converges to zero, one has

$\displaystyle \begin{aligned} \lim_{n\rightarrow\infty}|b_n\cdot c_n|\leq \lim_{n\rightarrow\infty}M\cdot |c_n|= M\cdot\lim_{n\rightarrow\infty}|c_n|=M\cdot 0=0. {} \end{aligned} $

(5.1.28)

Please note that for the final result, the vanishing limit of *c*
_{n}, i.e., in our case of the expression: has proved very important. If the sequence *c*
_{n} converged to some *g* ≠ 0, two subsequences of *a*
_{n} could be created: one composed of terms with even indexes (*a*
_{2k}) and the other with those of odd indexes (*a*
_{2k+1}), of which the former goes to *g*, and the latter to − *g*. This would mean that the sequence *a*
_{n} has no limit. This situation would be the case, if, for example, in the formula (5.1.24) we had cosine, instead of sine, i.e., if Proceeding in the similar way as above, one would come to It is easy to show that $\displaystyle \lim _{n\rightarrow \infty }\cos \left (\pi \sqrt {a^2+n^2}-n\pi \right )=1$. The sequence $a^{\prime }_n$ oscillates (it has two, the so-called “cluster” points: + 1 and − 1) and naturally has no limit. We are going to come back to this issue in Sect. 5.5 in Problems 1 and 2.

$\displaystyle \begin{aligned}\sin\left(\pi\sqrt{a^2+n^2}-n\pi\right) ,\end{aligned}$

$\displaystyle \begin{aligned} a^{\prime}_n=\cos\left(\pi\sqrt{a^2+n^2}\right). {} \end{aligned} $

(5.1.29)

$\displaystyle \begin{aligned} a^{\prime}_n=(-1)^n\cos\left(\pi\sqrt{a^2+n^2}-n\pi\right). {} \end{aligned} $

(5.1.30)

The convergence of the sequence will be proved and its limit will be found.

$\displaystyle \begin{aligned} a_n=n\left(\frac{\pi}{2}-\arccos\frac{1}{n}\right) {} \end{aligned} $

(5.1.31)

The general term is a product of two factors: The latter runs to zero since, due to the continuity of inverse cosine function, one can write The limit (5.1.31) is, therefore, of the type *∞*⋅ 0. To determine its value, one needs to know how fast the expression in brackets decreases to zero. One of the possible methods of proceeding in such circumstances is to examine the function $\arccos x$ for *x* → 0^{+}, e.g., by using its expansion in the Taylor series. We will, however, use another method, consisting of getting rid of the function that causes trouble—the cyclometric function. To achieve this goal we write As we know, the right-hand side of this equation is convergent to zero, hence also $a_n/n\underset {n\rightarrow \infty }{\longrightarrow }0 $. Now let us calculate the sine of both sides of (5.1.33). Why do we choose sine and not cosine, if in the formula one has arccos? Well, it is because on the right-hand side there is an additional *π*∕2, which, after the application of the reduction formula, will change the function sine into cosine: From this, one can compute 1∕*a*
_{n} and find the limit: where the known fact that has been exploited for *ϕ*, expressed in radian measure. Since the latter limit exists and is equal to 1, so, in accordance with Heine’s definition of the limit of a function (see (7.0.1)), the same result must hold for each sequence of arguments *ϕ*
_{n} convergent to zero. In particular, one can choose *ϕ*
_{n} = *a*
_{n}∕*n* obtaining (5.1.35).

$\displaystyle \begin{aligned} n\;\;\;\; \mathrm{and}\;\;\;\; \left(\frac{\pi}{2}-\arccos\frac{1}{n}\right). \end{aligned}$

$\displaystyle \begin{aligned} \lim_{n\rightarrow \infty}\arccos\frac{1}{n}= \arccos\lim_{n\rightarrow \infty}\frac{1}{n}=\arccos 0=\frac{\pi}{2}. {} \end{aligned} $

(5.1.32)

$\displaystyle \begin{aligned} \frac{a_n}{n}=\frac{\pi}{2}-\arccos\frac{1}{n}. {} \end{aligned} $

(5.1.33)

$\displaystyle \begin{aligned} \sin\frac{a_n}{n}=\sin\left(\frac{\pi}{2}-\arccos\frac{1}{n}\right)=\cos\left(\arccos\frac{1}{n}\right)=\frac{1}{n}=\frac{a_n}{n}\cdot \frac{1}{a_n}. {} \end{aligned} $

(5.1.34)

$\displaystyle \begin{aligned} \frac{1}{a_n}=\frac{\sin (a_n/n)}{a_n/n}\underset{n\rightarrow \infty}{\longrightarrow }1\;\implies\; a_n\underset{n\rightarrow \infty}{\longrightarrow }1 , {} \end{aligned} $

(5.1.35)

$\displaystyle \begin{aligned} \lim_{\phi\rightarrow 0}\frac{\sin\phi}{\phi}=1 {} \end{aligned} $

(5.1.36)

But how does one know the result (5.1.36) is true? The easiest way to justify it is to use Fig. 5.1, performed for 0 < *ϕ* < *π*∕2.

It shows a circle of radius 1 and two rectangular triangles: *AOB* and *COD* with the common angle *ϕ*. From △ *AOB* one has Similarly, from △ *COD* we obtain The length of the arc *CB* is just (in this measure) the value of the angle *ϕ*. One, therefore, has which allows us to write the following inequalities: Combining these will result in the form $\cos \phi <\sin \phi /\phi <1$. Because the functions $\cos \phi $ and $\sin \phi / \phi $ are even, this formula is valid also for − *π*∕2 < *ϕ* < 0. Taking now an arbitrary sequence of arguments $\phi _n \underset {n\rightarrow \infty }{\longrightarrow } 0$, one can apply the squeeze theorem, one is already familiar with: and the desired conclusion (5.1.36) is reached. This result will turn out very useful in many different problems and it is worth remembering.

$\displaystyle \begin{aligned} \sin\phi=\frac{|AB|}{|OB|}=\frac{|AB|}{1}=|AB|. {} \end{aligned} $

(5.1.37)

$\displaystyle \begin{aligned} \tan\phi=\frac{|CD|}{|OC|}=\frac{|CD|}{1}=|CD|. {} \end{aligned} $

(5.1.38)

$\displaystyle \begin{aligned} \sin\phi=|AB|<\phi<|CD|=\tan\phi , {} \end{aligned} $

(5.1.39)

$\displaystyle \begin{aligned} \sin\phi<\phi ,\;\;\;\;\mathrm{and}\;\;\;\; \tan\phi>\phi\;\;\implies\;\;\sin\phi>\phi\cos\phi. {} \end{aligned} $

(5.1.40)

$\displaystyle \begin{aligned} 1\underset{n\rightarrow \infty}{\longleftarrow}\cos\phi_n<\frac{\sin\phi_n}{\phi_n}<1\underset{n\rightarrow \infty}{\longrightarrow}1 , {} \end{aligned} $

(5.1.41)

Using the definition of the limit, the convergence of the sequence will be examined, depending on the values of the parameters $l,k\in \mathbb {N}$, where *p*
_{k}(*x*) = *α*
_{0} + *α*
_{1}*x* + … + *α*
_{k}*x*
^{k} and *q*
_{l}(*x*) = *β*
_{0} + *β*
_{1}*x* + … + *β*
_{l}*x*
^{l} are real polynomials. It is assumed that at least *α*
_{k} and *β*
_{l} do not vanish. In the case of convergence, the limit of *a*
_{n} will be found.

$\displaystyle \begin{aligned} a_n=\frac{p_k(n)}{q_l(n)} {} \end{aligned} $

(5.2.1)

It is worth solving one example referring directly to the definition of the limit. For this goal, we chose the above limit; a step that appears as a component in many different problems and one absolutely must know it. Estimations, performed below, are not quite obvious. Let us start with the reminder of the definition of the limit: However, it should be emphasized that if one wishes to use this definition in the proof, one must first be able to guess the value of *g*.

$\displaystyle \begin{aligned} \lim_{n\rightarrow\infty}a_n=g\;\iff\; \forall_{\epsilon>0}\;\; \exists_{N\in\mathbb{N}}\;\;\forall_{n\geq N}\;\;\;\;\; |a_n-g|<\epsilon. {} \end{aligned} $

(5.2.2)

Let us now write (5.2.1) in the form

$\displaystyle \begin{aligned} \begin{array}{rcl} a_n&\displaystyle =&\displaystyle \frac{p_k(n)}{q_l(n)}=\frac{\alpha_0+\alpha_1n+\ldots+\alpha_{k-1}n^{k-1}+\alpha_kn^k}{\beta_0+\beta_1n+\ldots+\beta_{l-1}n^{l-1}+\beta_ln^l}\\ &\displaystyle =&\displaystyle \frac{n^k}{n^l}\cdot \frac{\alpha_0/n^k+\alpha_1/n^{k-1}+\ldots+\alpha_{k-1}/n+\alpha_k}{\beta_0/n^l+\beta_1/n^{l-1}+\ldots+\beta_{l-1}/n+\beta_l}. {} \end{array} \end{aligned} $

(5.2.3)

The sequence will behave differently depending on which polynomial (that in the numerator or that in the denominator) has a higher degree, or whether the degrees are identical. Therefore, we are going to proceed with three independent cases

$\displaystyle \begin{aligned} l>k , \; \; \; \; l<k, \; \; \; \mathrm{and} \; \; \; \; l=k. {} \end{aligned} $

(5.2.4)

Case 1: *l* > *k*

Due to the fact that *l* − *k* is positive, by choosing a sufficiently large value of *n*, one will certainly be able to make the factor *n*
^{k}∕*n*
^{l} = 1∕*n*
^{l−k} as small as one wishes. So now one has to focus on evaluating the second factor in (5.2.3). With the numerator, the matter is simple: since *n* ≥ 1, then For the denominator, we need a reverse estimate—from below. It is not evident at all and one has to ponder over the expression It is intuitively felt that for suitably large values of *n*, the first *l* terms of the sum will be small (as modules) compared to the last component, i.e., *β*
_{l}. Let us use this intuition to find an appropriate estimate. One simply has to determine how large *n* should be. For instance we might like the sum of the former *l* terms to be smaller than, say, a half of the last one, i.e., |*β*
_{l}|∕2. Then it could be written that the denominator is estimated from below as follows: Because the number of terms of the form *β*
_{i}∕*n*
^{l−i}, where *i* = 0, 1, …, *l* − 1, is equal to *l*, let us require that each of them must be less (again as a modulus) than $\left | 1/l\cdot \beta _l/2\right | $. Then their sum certainly does not exceed the value |*β*
_{l}|∕2.

$\displaystyle \begin{aligned} \left|\frac{\alpha_0}{n^k}+\frac{\alpha_1}{n^{k-1}}+\ldots+\frac{\alpha_{k-1}}{n}+\alpha_k\right|\leq |\alpha_0|+|\alpha_1|+\ldots+|\alpha_{k-1}|+|\alpha_k|. {} \end{aligned} $

(5.2.5)

$\displaystyle \begin{aligned}\frac{\beta_0}{n^l}+\frac{\beta_1}{n^{l-1}}+\ldots+\frac{\beta_{l-1}}{n}+\beta_l.\end{aligned} $

$\displaystyle \begin{aligned} \left|\frac{\beta_0}{n^l}+\frac{\beta_1}{n^{l-1}}+\ldots+\frac{\beta_{l-1}}{n}+\beta_l\right|> |\beta_l|-\frac{1}{2}|\beta_l|=\frac{1}{2}|\beta_l|. {} \end{aligned} $

(5.2.6)

What *n* should be chosen for this purpose? Well, such that, for all values of *i*, Let us remember this restriction. We know from the text of the exercise that *β*
_{l} ≠ 0, so this expression is plausible. With the above assumption for *n*, one has: For large *n*, this expression becomes arbitrarily small for any positive constant *C*. It is anticipated, therefore, that this limit of the sequence is 0. Hence, if we choose, in accordance with (5.2.2), any small *𝜖* > 0, it ought to be shown that there exists such $N\in \mathbb {N} $, that for *n* ≥ *N* one has When looking at (5.2.8), it is seen that this requirement could really be satisfied, by taking Recalling the conditions (5.2.7) formulated earlier, we choose as *N* (which is needed by the definition of a limit) any natural number satisfying With this choice, (5.2.9) is actually fulfilled and the proof is complete.

$\displaystyle \begin{aligned} \left|\frac{\beta_i}{n^{l-i}}\right|<\frac{1}{2l}|\beta_l|\;\iff\; n>\sqrt[l-i]{2l\left|\beta_i/\beta_l\right|}. {} \end{aligned} $

(5.2.7)

$\displaystyle \begin{aligned} |a_n|<\frac{1}{n^{l-k}}\cdot\frac{|\alpha_0|+|\alpha_1|+\ldots+|\alpha_{k-1}|+|\alpha_k|}{\frac{1}{2}|\beta_l|}=:C\;\frac{1}{n^{l-k}}. {} \end{aligned} $

(5.2.8)

$\displaystyle \begin{aligned} |a_n-0| = |a_n| < \epsilon. {}\end{aligned} $

(5.2.9)

$\displaystyle \begin{aligned} n>\sqrt[l-k]{C/\epsilon}. {} \end{aligned} $

(5.2.10)

$\displaystyle \begin{aligned} \begin{array}{rcl} N &\displaystyle >&\displaystyle \max\bigg\{\sqrt[l]{2l\left|\beta_0/\beta_l\right|}, \sqrt[l-1]{2l\left|\beta_1/\beta_l\right|},\\ &\displaystyle \ldots&\displaystyle , \sqrt{2l\left|\beta_{l-2}/\beta_l\right|}, 2l\left|\beta_{l-1}/\beta_l\right|, \sqrt[l-k]{C/\epsilon}\bigg\}. {} \end{array} \end{aligned} $

(5.2.11)

Case 2: *l* < *k*

Now, one will need inverse estimates of the numerator and the denominator in (5.2.3). However, the transformations performed in case 1 can be reused, swapping the roles of the numerator and the denominator. So, we have the counterpart of the formula (5.2.5): and of the formula (5.2.6): as long as for all *i* = 0, 1, …, *k* − 1. Taken together, this result means: Since *D* > 0, for any positive constant *M*, using we obtain |*a*
_{n}| > *M*. Almost all terms of the sequence |*a*
_{n}| are greater than the arbitrary constant! This sequence cannot, therefore, be bounded. It must be divergent, and the same must refer to the sequence *a*
_{n} itself.

$\displaystyle \begin{aligned} \left|\frac{\beta_0}{n^l}+\frac{\beta_1}{n^{l-1}}+\ldots+\frac{\beta_{l-1}}{n}+\beta_l\right|\leq |\beta_0|+|\beta_1|+\ldots+|\beta_{l-1}|+|\beta_l| {} \end{aligned} $

(5.2.12)

$\displaystyle \begin{aligned} \left|\frac{\alpha_0}{n^k}+\frac{\alpha_1}{n^{k-1}}+\ldots+\frac{\alpha_{k-1}}{n}+\alpha_k\right|> |\alpha_k|-\frac{1}{2}|\alpha_k|=\frac{1}{2}|\alpha_k| , {} \end{aligned} $

(5.2.13)

$\displaystyle \begin{aligned} n>\sqrt[k-i]{2k\left|\alpha_i/\alpha_k\right|}, {} \end{aligned} $

(5.2.14)

$\displaystyle \begin{aligned} |a_n|>n^{k-l}\cdot\frac{|\alpha_k|/2}{|\beta_0|+|\beta_1|+\ldots+|\beta_{k-1}|+|\beta_k|}=:D\cdot n^{k-l}. {} \end{aligned} $

(5.2.15)

$\displaystyle \begin{aligned} \begin{array}{rcl} n &\displaystyle >&\displaystyle \max\bigg\{\sqrt[k]{2k\left|\alpha_0/\alpha_k\right|}, \sqrt[k-1]{2k\left|\alpha_1/\alpha_k\right|},\\ &\displaystyle \ldots&\displaystyle , \sqrt{2k\left|\alpha_{k-2}/\alpha_k\right|}, 2k\left|\alpha_{k-1}/\alpha_k\right|, \sqrt[k-l]{M/D}\bigg\} , {} \end{array} \end{aligned} $

(5.2.16)

Case 3: *l* = *k*

Our purpose will be to demonstrate that the limit constitutes the number *g* = *α*
_{k}∕*β*
_{k}. Therefore, let us consider the difference *a*
_{n} − *g*, i.e., One has managed to have the same issue as the first case: the polynomial in the numerator has a lower degree than that in the denominator! We already know that by choosing a sufficiently large *N* (for any previously indicated *𝜖*), the inequality can be satisfied. The indicated number is actually the limit. This result should be memorized: if the degrees of polynomials in the numerator and the denominator are equal, the limit of the sequence is the quotient of the coefficients accompanying the highest powers of *n*.

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \frac{p_k(n)}{q_k(n)}- \frac{\alpha_k}{\beta_k}= \frac{\alpha_0+\alpha_1n+\ldots+\alpha_{k-1}n^{k-1}+\alpha_kn^k}{\beta_0+\beta_1n+\ldots+\beta_{k-1}n^{k-1}+\beta_kn^k}-\frac{\alpha_k}{\beta_k}{}\\ &\displaystyle &\displaystyle =\frac{(\alpha_0\beta_k-\alpha_k\beta_0)+(\alpha_1\beta_k-\alpha_k\beta_1)n+\ldots+(\alpha_{k-1}\beta_k-\alpha_k\beta_{k-1})n^{k-1}}{\beta_k(\beta_0+\beta_1n+\ldots+\beta_{k-1}n^{k-1}+\beta_kn^k)}. \end{array} \end{aligned} $

(5.2.17)

$\displaystyle \begin{aligned} \left|a_n-\frac{\alpha_k}{\beta_k}\right|<\epsilon {} \end{aligned} $

(5.2.18)

Applying Cauchy’s test, the convergence of the sequence will be examined.

$\displaystyle \begin{aligned} a_n=\frac{n\left(1+1/n\right)^{n^2}}{n^2+n2^n+3^n} {} \end{aligned} $

(5.2.19)

In accordance with the text of the problem, we are going to solve it using Cauchy’s criterion called also the “*n*th root test.” As we know, in this case, one needs to calculate $\sqrt [n]{|a_n|}$ and then examine the limit of this expression for *n* →*∞*, and if One can ask the question, how does one know that this criterion is the most convenient to study the expression (5.2.19). Well, it is suggested by its structure in which the most “troublesome” part seems to be the following factor in the numerator: In the absence of Cauchy’s criterion, one should find some nontrivial estimate for it. It should be noted that the other expressions in (5.2.19) are relatively easy to estimate from above or below if necessary. Guided by the principle of adapting the criterion to the most difficult part of an expression, we choose Cauchy’s criterion, for which the *n*th root appears and For now, we do not care about the other elements of the formula (5.2.19) because one can see that we will certainly handle them. One has $\sqrt [n]{n}\underset {n\rightarrow \infty }{\longrightarrow }1$ and with expressions similar to that in the denominator we learned how to deal with solving Problem 3 in Sect. 5.1.

$\displaystyle \lim _{n\rightarrow \infty }\sqrt [n]{|a_n|}<1$, then lim

_{n→∞}*a*_{n}= 0.$\displaystyle \lim _{n\rightarrow \infty }\sqrt [n]{|a_n|}>1$, then the sequence is divergent.

$\displaystyle \lim _{n\rightarrow \infty }\sqrt [n]{|a_n|}=1$, this criterion does not determine convergence and the sequence ought to be examined with other methods.

$\displaystyle \begin{aligned}\left(1+\frac{1}{n}\right)^{n^2}.\end{aligned}$

$\displaystyle \begin{aligned} \sqrt[n]{\left(1+\frac{1}{n}\right)^{n^2}}=\left(1+\frac{1}{n}\right)^n\underset{n\rightarrow\infty}{\longrightarrow} e. {} \end{aligned} $

(5.2.20)

The above considerations constitute only a motivation for the choice of this and not of the other criterion. Now it is time to apply them to the relevant calculations. One finds

$\displaystyle \begin{aligned} \sqrt[n]{|a_n|}=\sqrt[n]{\frac{n\left(1+1/n\right)^{n^2}}{n^2+n2^n+3^n}}=\frac{\sqrt[n]{n}\left(1+1/n\right)^n}{\sqrt[n]{n^2+n2^n+3^n}}. {} \end{aligned} $

(5.2.21)

We would like now to make use of the rule that the limit of the product (or quotient) is equal to the product (quotient) of limits, in so far as they all exist. Sequences in the numerator have limits respectively equal to 1 and *e*, which has been discussed above. It only remains to verify the limit of the sequence in the denominator. For our transformations to make sense, it must be obviously different from zero. The leading expression is naturally 3^{n}, so for sufficiently large *n* one can estimate Thanks to the squeeze theorem, one gets Finally, one obtains and consequently

$\displaystyle \begin{aligned} \begin{array}{rcl} \sqrt[n]{n^2+n2^n+3^n}&\displaystyle <&\displaystyle \sqrt[n]{3^n+3^n+3^n}=\sqrt[n]{3\cdot 3^n}=3\sqrt[n]{3}\underset{n\rightarrow\infty}{\longrightarrow} 3\cdot 1=3 ,\\ \sqrt[n]{n^2+n2^n+3^n}&\displaystyle >&\displaystyle \sqrt[n]{0+0+3^n}=\sqrt[n]{3^n}=3\underset{n\rightarrow\infty}{\longrightarrow} 3. {} \end{array} \end{aligned} $

(5.2.22)

$\displaystyle \begin{aligned} \lim_{n\rightarrow\infty}\sqrt[n]{n^2+n2^n+3^n}=3. {} \end{aligned} $

(5.2.23)

$\displaystyle \begin{aligned} \lim_{n\rightarrow\infty}\sqrt[n]{|a_n|}=\frac{1\cdot e}{3}<1 , {} \end{aligned} $

(5.2.24)

$\displaystyle \begin{aligned} \lim_{n\rightarrow \infty}a_n=0. {} \end{aligned} $

(5.2.25)

At the beginning of our considerations, Cauchy’s criterion was recalled, but an attentive reader would notice that a particular situation, where $\displaystyle \lim _{n\rightarrow \infty }\sqrt [n]{|a_n|}$ simply does not exist, was omitted. We are going to come back to this issue (in the context of series) in Exercise 4 of Sect. 13.2.

Using d’Alembert’s criterion, the convergence of the sequence will be examined.

$\displaystyle \begin{aligned} a_n=\frac{(n/2)^n(n+1)}{(2n+1)!!(n+2)} {} \end{aligned} $

(5.2.26)

Let us first recall the d’Alembert criterion. First, one creates the quotient *a*
_{n+1}∕*a*
_{n} and looks for its limit as *n* →*∞*. Then, if As it is easy to understand, this criterion is particularly convenient if the formula for *a*
_{n} contains problematic factors that cancel or lead to the known limits when dividing *a*
_{n+1} by *a*
_{n}. In our example, such factors do exist: it is an expression with a factorial, so almost everything will be canceled in the quotient, and (*n*∕2)^{n} in the numerator will at most lead to the limit of the type (5.2.20). Other elements are completely irrelevant because their quotients will go to 1 and, therefore, do not affect the conclusions drawn from the d’Alembert criterion. Therefore, equally well, one could consider a sequence with the general term Coming back to (5.2.26), we calculate now

$\displaystyle \lim _{n\rightarrow \infty }\left |\frac {a_{n+1}}{a_n}\right |<1$, then lim

_{n→∞}*a*_{n}= 0.$\displaystyle \lim _{n\rightarrow \infty }\left |\frac {a_{n+1}}{a_n}\right |>1$, the sequence

*a*_{n}is divergent.$\displaystyle \lim _{n\rightarrow \infty }\left |\frac {a_{n+1}}{a_n}\right |=1$, this criterion does not determine convergence and the sequence ought to be examined with other methods.

$\displaystyle \begin{aligned} a_n=\frac{(n/2)^n}{(2n+1)!!}. {} \end{aligned} $

(5.2.27)

$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{a_{n+1}}{a_n}&\displaystyle =&\displaystyle \frac{((n+1)/2)^{n+1}(n+2)}{(2n+3)!!(n+3)}\left[\frac{(n/2)^n(n+1)}{(2n+1)!!(n+2)}\right]^{-1}\\ &\displaystyle =&\displaystyle \frac{(n+1)/2)^n}{(n/2)^n}\cdot\frac{(n+1)/2}{2n+3}\cdot\frac{(n+2)(n+2)}{(n+3)(n+1)}. {} \end{array} \end{aligned} $

(5.2.28)

Now let us have a look at all fractions. One has, therefore and hence, by virtue of the d’Alembert criterion, it can be deduced that lim_{n→∞}*a*
_{n} = 0.

The first one can be rewritten in the form $\left ((n+1)/n\right )^n=(1+1/n)^n$, so its limit is well known (and equal to

*e*).The second fraction has the limit of 1∕4, which we know already from the first problem in this section.

The last one, in accordance with what has already been stated, does not matter because it goes to 1.

$\displaystyle \begin{aligned} \lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|= e\cdot\frac{1}{4}\cdot 1<1\; {} \end{aligned} $

(5.2.29)

As one can see, the most important element is the ability to choose a suitable criterion to a specific problem. Nothing can supersede here the experience coming from solving multiple problems. It is then easier to recognize in the expression the important elements and those without any influence on the selection of the criterion. In this example, the key elements were undoubtedly factors present in (5.2.27), and certainly not a fraction (*n* + 1)∕(*n* + 2).

At the end of this solution, let us consider what may happen in case the expression $ \left |a_{n+1}/a_n\right |$ has no limit at all. If $\left |a_{n+1}/a_n\right |\underset {n\rightarrow \infty }{\longrightarrow }\infty $, then for an arbitrarily large number *M* and for almost all *n* we have $\left |a_{n+1}/a_n\right | >M$. The sequence |*a*
_{n}| behaves “worse” than a geometric sequence of quotient *M*. |*a*
_{n}| (and, therefore, also *a*
_{n}) must be divergent. However, if $\left |a_{n+1 }/a_n\right |$ has no limit, but still does not tend to infinity, then even if $\limsup _{n\rightarrow \infty }\left |a_{n+1}/a_n\right |=q>1$, the sequence *a*
_{n} can be convergent (even for *q* = *∞*), which can be found out while considering the example of *b*
_{n} with the following consecutive terms: The odd and even subsequences are independent geometric sequences with quotient equal to 2∕3, and, therefore, both tend to zero, while we have The certainty of the convergence of *a*
_{n} to 0 takes place, however, only when

$\displaystyle \begin{aligned} 1,\;\frac{1}{3},\;\frac{2}{3},\;\frac{2}{9},\;\frac{4}{9},\;\frac{4}{27},\;\frac{8}{27},\;\ldots . {} \end{aligned} $

(5.2.30)

$\displaystyle \begin{aligned} \limsup_{n\rightarrow\infty}\left|\frac{b_{n+1}}{b_n}\right|=2 ,\;\;\;\; \liminf_{n\rightarrow\infty}\left|\frac{b_{n+1}}{b_n}\right|=\frac{1}{3}. {} \end{aligned} $

(5.2.31)

$\displaystyle \begin{aligned}\limsup_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|<1.\end{aligned}$

Using the Stolz–Cesàro criterion, the convergence of the sequence will be examined.

$\displaystyle \begin{aligned} a_n=\frac{1^{1/4} + 3^{1/4} +\ldots + (2n+1)^{1/4}}{n^{5/4}} {} \end{aligned} $

(5.2.32)

The Stolz–Cesàro criterion (or theorem) allows us to find the limits of sequences of the type when both *b*
_{n} and *c*
_{n} go to infinity, and the sequence *c*
_{n} is monotonically increasing. This criterion tells us that, rather than examining the former limit, one can instead look for and both these limits shall be equal. Interestingly, this criterion is also applicable when $a_n= b_n/c_n\underset {n\rightarrow \infty }{\longrightarrow }\infty $.

$\displaystyle \begin{aligned}a_n=\frac{b_n}{c_n} ,\end{aligned}$

$\displaystyle \begin{aligned} \lim_{n\rightarrow\infty}\frac{b_{n+1}-b_n}{c_{n+1}-c_n}, {} \end{aligned} $

(5.2.33)

There appears a quite natural question: why should it be more comfortable studying the behavior of the fraction (*b*
_{n+1} − *b*
_{n})∕(*c*
_{n+1} − *c*
_{n}) than the original expression *b*
_{n}∕*c*
_{n}? Well, there are many examples of sequences that contain sums of many terms in the numerator or denominator. Creating a difference of the type *b*
_{n+1} − *b*
_{n}, we hope that a large number of them will reduce, and then a simpler expression remains. This is just the case for our problem. Additionally, one can also hope that degrees of divergences in the numerator and the denominator will decrease. In the present example, Creating a difference of expressions in the numerator, one gets and in the denominator In this way, our expression has been significantly simplified and reduced (5.2.32) to the limit A potential problem here still can appear in the denominator in which the difference of two expressions diverging to infinity is present. However, we have already learned how to deal with this type of situation in the Example 1 of Sect. 5.1. Following the idea elaborated there, we write Using now the formula one sees that terms with *n*
^{5} in the denominator are canceled. Extracting the highest powers of *n* from the numerator and from the denominator, one gets and this is also the limit of the sequence *a*
_{n}.

$\displaystyle \begin{aligned} \begin{array}{rcl} b_n &\displaystyle =&\displaystyle 1^{1/4} + 3^{1/4} +\ldots + (2n+1)^{1/4} ,\\ c_n &\displaystyle =&\displaystyle n^{5/4} . {} \end{array} \end{aligned} $

(5.2.34)

$\displaystyle \begin{aligned} \begin{array}{rcl} b_{n+1}-b_n &\displaystyle =&\displaystyle 1^{1/4} + 3^{1/4} +\ldots + (2n+1)^{1/4}+ (2n+3)^{1/4}\\ &\displaystyle -&\displaystyle 1^{1/4} - 3^{1/4} -\ldots - (2n+1)^{1/4}=(2n+3)^{1/4} ,\\ {} \end{array} \end{aligned} $

(5.2.35)

$\displaystyle \begin{aligned} c_{n+1}-c_n = (n+1)^{5/4} - n^{5/4}. {}\end{aligned} $

(5.2.36)

$\displaystyle \begin{aligned} \lim_{n\rightarrow\infty}\frac{(2n+3)^{1/4}}{(n+1)^{5/4} - n^{5/4}}. {}\end{aligned} $

(5.2.37)

$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{(2n+3)^{1/4}}{(n+1)^{5/4} - n^{5/4}}&\displaystyle =&\displaystyle \frac{(2n+3)^{1/4}}{(n+1)^{5/4} - n^{5/4}}\cdot \frac{(n+1)^{5/4} + n^{5/4}}{(n+1)^{5/4} + n^{5/4}{}}\\ &\displaystyle =&\displaystyle \frac{(2n+3)^{1/4}((n+1)^{5/4} + n^{5/4})}{(n+1)^{5/2} - n^{5/2}}\\ &\displaystyle =&\displaystyle \frac{(2n+3)^{1/4}((n+1)^{5/4} + n^{5/4})}{(n+1)^{5/2} - n^{5/2}}\cdot \frac{(n+1)^{5/2} + n^{5/2}}{(n+1)^{5/2} + n^{5/2}}\\ &\displaystyle =&\displaystyle \frac{(2n+3)^{1/4}((n+1)^{5/4} + n^{5/4})((n+1)^{5/2} + n^{5/2})}{(n+1)^5 - n^5}. \end{array} \end{aligned} $

(5.2.38)

$\displaystyle \begin{aligned} (n+1)^5 =n^5 +5n^4 +10n^3 +10n^2 +5n+1 ,\end{aligned} $

$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \lim_{n\rightarrow\infty}\frac{b_{n+1}-b_n}{c_{n+1}-c_n}={}\\ &\displaystyle &\displaystyle =\lim_{n\rightarrow\infty}\frac{n^4\left(2+3/n\right)^{1/4}\left(\left(1+1/n\right)^{5/4} + 1\right)\left(\left(1+1/n\right)^{5/2} + 1\right)}{n^4\left(5+10/n+10/n^2+5/n^3+1/n^4\right)} =\frac{2^{9/4}}{5}, \end{array} \end{aligned} $

(5.2.39)

The limit of the sequence will be found.

$\displaystyle \begin{aligned} a_n=\left(\frac{n^3+100\sin n^2+1}{n^3+n^2+\log^5 n}\right)^{n^2/(n+1)} {}\end{aligned} $

(5.2.40)

When setting about solving similar problems, one should not be subject to the false impression that it is too complicated or even unsolvable. As will be seen below, one needs only to know how to separate, what is important from, what is insignificant, and the example turns out to be very simple. But first one must recall the theorem that will be used. We know from the lecture of analysis that if a sequence *a*
_{n} is of the special form where *b*
_{n} has a vanishing limit and *c*
_{n} behaves so, that then the limit of *a*
_{n} itself may be easily found as:

$\displaystyle \begin{aligned} a_n=(1+b_n)^{c_n} , {}\end{aligned} $

(5.2.41)

$\displaystyle \begin{aligned} \lim_{n\rightarrow\infty} b_nc_n=g\neq\pm\infty , {}\end{aligned} $

(5.2.42)

$\displaystyle \begin{aligned} \lim_{n\rightarrow \infty} a_n=e^g. {} \end{aligned} $

(5.2.43)

If one now looks at our expression (5.2.40), it will easily noticed that it has the desired character. A fraction in brackets, despite its complex form, in an obvious way goes to 1, due to the fact that the numerator and denominator have identical leading terms (i.e., *n*
^{3}). This unity may be separated in order to reach the structure such as in (5.2.41). The exponent, in turn, diverges to infinity, which gives hope to meet (5.2.42). Hence, let us assume that:

$\displaystyle \begin{aligned} b_n=\frac{n^3+100\sin n^2+1}{n^3+n^2+\log^5 n}-1 ,\;\;\;\;\; c_n=\frac{n^2}{n+1}. {} \end{aligned} $

(5.2.44)

First, let us concentrate on *b*
_{n}. In the numerator we recognize that in addition to *n*
^{3}, which runs to infinity, other expressions are bounded. The sine expression is insignificant even with a large amplitude since the whole numerator can be estimated as *n*
^{3} ± *M*, where *M* is a constant. For the limit of *a*
_{n}, this constant is completely irrelevant, and hence one could write in the numerator *n*
^{3} ± 1 or even simply *n*
^{3}. What could eventually prove to be important is a term behaving as *n*
^{2} or worse, but there is no such term in the numerator, unlike the denominator.

So, let us look at the denominator. If it is rewritten in the form it can be seen that the last term decreases very quickly to zero as compared with the first ones, so again, one could skip it and leave the denominator in the simplified form of $n^3\left (1 +1/n\right )$.

$\displaystyle \begin{aligned} n^3+n^2+\log^5 n=n^3\left(1+\frac{1}{n}+\frac{\log^5n}{n^3}\right) , {} \end{aligned} $

(5.2.45)

Now the question may arise: how do we know that terms such as *n*
^{2} (or worse) appearing with *n*
^{3} are important, while others can be omitted with impunity? Well, the answer is given by the form of the sequence *c*
_{n} together with the condition (5.2.42). For large *n* one has *c*
_{n} ≃ *n*, and a finite limit *g* in (5.2.42) would be obtained when *b*
_{n} ≃ 1∕*n*. Since both leading expressions in the numerator and the denominator are of the form *n*
^{3}, these giving (at the end) finite limit are *n*
^{3} ⋅ 1∕*n* = *n*
^{2}. In the possible case of vanishing of such terms—which is not the case here—one eventually has to take into account further ones.

So, the following conclusion occurs: instead of calculating the limit of the complicated sequence *a*
_{n}, we are going to look for the limit of since they are identical. And the latter may be obtained almost mentally as equal to *e*
^{−1} = 1∕*e*, because

$\displaystyle \begin{aligned} \left(\frac{n^3}{n^3+n^2}\right)^{n^2/(n+1)}=\left(\frac{n^3+n^2-n^2}{n^3+n^2}\right)^{n^2/(n+1)}=\left(1-\frac{1}{n+1}\right)^{n^2/(n+1)} , {} \end{aligned} $

(5.2.46)

$\displaystyle \begin{aligned} g=\lim_{n\rightarrow\infty}\frac{-1}{n+1}\cdot \frac{n^2}{n+1}=-1. {} \end{aligned} $

(5.2.47)

Let us now return to the full formula for *a*
_{n} because the replacement of the expression (5.2.40) with (5.2.46) was based only on our intuition and not on strict arguments. One has which is in accordance with our predictions. Expressions with sine or logarithm have proved to be only “decorations” which do not have any impact on the limit. Thus, definitively

$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{n\rightarrow\infty} b_n\,c_n&\displaystyle =&\displaystyle \lim_{n\rightarrow\infty}\left(\frac{n^3+100\sin n^2+1}{n^3+n^2+\log^5 n}-1\right)\,\frac{n^2}{n+1}{}\\ &\displaystyle =&\displaystyle \lim_{n\rightarrow\infty}\frac{-n^2-\log^5 n+100\sin n^2+1}{n^3+n^2+\log^5 n}\cdot\frac{n^2}{n+1}\\ &\displaystyle =&\displaystyle \lim_{n\rightarrow\infty}\frac{n^4}{n^4}\cdot\frac{-1-1/n^2\cdot\log^5 n+100\sin n^2/n^2+1/n^2}{(1+1/n+1/n^3\cdot\log^5 n)(1+1/n)}=-1 , \end{array} \end{aligned} $

(5.2.48)

$\displaystyle \begin{aligned} \lim_{n\rightarrow \infty} a_n=e^{-1}=\frac{1}{e}. {} \end{aligned} $

(5.2.49)

The limit of the sequence defined by the recursive formula for *n* = 0, 1, 2, …, will be calculated, where the two initial members are chosen as *a*
_{0} = 5 and *a*
_{1} = −1∕2.

$\displaystyle \begin{aligned} a_{n+2}=\frac{1}{4}\, a_{n+1}+\frac{1}{8}\, a_n, {} \end{aligned} $

(5.3.1)

The sequence (5.3.1) is defined by a linear recursive equation, which is homogeneous (i.e., without free terms), and the factors multiplying *a*
_{k} are constant numbers. In such a case, we have at our disposal a method, thanks to which not only a limit can be found (in so far as it exists), but also a general formula for an arbitrary sequence member. The only significant complication, which could eventually arise, is related to the possible high degree of the recursive equation to which we are going to return to later.

The idea is the following: let us substitute for the sequence terms *a*
_{n} = *λ*
^{n}, where *λ* is a certain number other than zero. Now it should be verified if there exists such value of *λ*, for which the equation is satisfied. For *a*
_{n} chosen in that way, the increase of *n* simply means multiplying by a constant, and this is, essentially, the sense of the equation (5.3.1).

After the above substitution and simplifying both sides, one gets the quadratic equation which has two solutions: *λ*
_{1} = 1∕2 and *λ*
_{2} = −1∕4. “Forgetting” for a moment that one has to comply with initial conditions, there are two possibilities:

$\displaystyle \begin{aligned} \lambda^2-\frac{1}{4}\;\lambda-\frac{1}{8}=0 , {} \end{aligned} $

(5.3.2)

$\displaystyle \begin{aligned} a^{\prime}_n=\left(\frac{1}{2}\right)^n\;\;\;\; \mathrm{and}\;\;\;\; a^{\prime\prime}_n=\left(-\frac{1}{4}\right)^n. {} \end{aligned} $

(5.3.3)

An important feature of linear and homogeneous equations is that any solution multiplied by a constant is also a solution. And any two solutions added to each other form a solution too. This means that the most general formula for *a*
_{n} should take the form where *α* and *β* are constants. They can be fixed with the use of initial conditions. Substituting *n* = 0, one has *α* + *β* = 5, and substituting *n* = 1, one finds 1∕2 ⋅ *α* − 1∕4 ⋅ *β* = −1∕2. Thus, *α* = 1 and *β* = 4 and the complete solution is given by the expression Since in both terms the bases are smaller than 1, the limit of *a*
_{n} for *n* →*∞* is equal to 0.

$\displaystyle \begin{aligned} a_n=\alpha\left(\frac{1}{2}\right)^n+\beta \left(-\frac{1}{4}\right)^n , {} \end{aligned} $

(5.3.4)

$\displaystyle \begin{aligned} a_n=\left(\frac{1}{2}\right)^n+4 \left(-\frac{1}{4}\right)^n. {} \end{aligned} $

(5.3.5)

*The equation for a constant**λ*,*similar to*(5.3.2)*, may have a double root**λ*=*λ*_{0}.In this case, one of the solutions has still the form $a^{\prime }_n= \lambda _0^n$, but apparently the second solution is missing. In general, in order to meet the initial conditions, we must have at our disposal two constants*α*and*β*, as in (5.3.4). This, however, means that one needs*two*solutions. Substituting expression $a^{\prime \prime }_n=n\lambda _0^n$ into the recursive equation, which then would have to be in the form $a_{n+2}=2\lambda _0a_{n+1}-\lambda _0^2 a_n$ (only in this case*λ*_{0}is a double root), one can easily see that this equation is in fact satisfied. So, in place of (5.3.4), we have now the formulaand further reasoning remains without changes.$\displaystyle \begin{aligned} a_n=\alpha\lambda_0^n+\beta n\lambda_0^n , {} \end{aligned} $(5.3.6)*The equation for a constant**λ**may have complex roots.*One need not worry about complex values of lambdas. All calculations can be carried out exactly in the same way, and if all the coefficients in the recursive equation and the initial constants were real, then the final result surely can be written as real. For, starting from two real constants

*a*_{0}and*a*_{1}and applying only addition and multiplication by real parameters, as in (5.3.1), one can never come to any complex numbers.*The equation for a constant**λ**may be of high degree.*If the recurrence is not “by 2,” but for example “by

*k*,” we face the necessity of solving, instead of (5.3.2), an algebraic equation of respectively high order (equal to the degree of the recursive equation). This task can be very difficult and constitutes a separate complex issue. However, if one manages to find all roots, the general solution can be written in the form of (5.3.4) or (5.3.6), in which there will now be*k*terms and*k*arbitrary constants. The same will be the number of initial conditions, so all constants will be determined.

$\displaystyle \begin{aligned} v_n=\left[\begin{array}{c} a_{n+1}\\ a_{n} \end{array}\right]. \end{aligned} $

(5.3.7)

$\displaystyle \begin{aligned} M=\left[\begin{array}{cc} 1/4 & 1/8 \\ 1 & 0 \end{array}\right]. \end{aligned} $

(5.3.8)

$\displaystyle \begin{aligned} v_0=\left[\begin{array}{c} -1/2\\ 5 \end{array}\right]. \end{aligned} $

(5.3.9)

$\displaystyle \begin{aligned}\phi(\lambda)=\lambda^2-\frac{1}{4}\;\lambda-\frac{1}{8}\end{aligned}$

If one denotes the eigenvectors with symbols *u*
_{1} and *u*
_{2}, the subsequent steps of the solution are already clear from the course of algebra: *v*
_{0} is expanded in the basis of eigenvectors *v*
_{0} = *c*
_{1}*u*
_{1} + *c*
_{2}*u*
_{2} (constants *c*
_{1,2} are then known) and it can be written as The solution (5.3.5) can now be read off from (5.3.10), comparing the lower components of the vector *v*
_{n} on both sides.

$\displaystyle \begin{aligned} \begin{array}{rcl} {} v_n=M^nv_0&\displaystyle =&\displaystyle M^n(c_1u_1+c_2u_2)=c_1M^nu_1+c_2M^nu_2\\ &\displaystyle =&\displaystyle c_1(\lambda_1)^nu_1+c_2(\lambda_2)^nu_2. \end{array} \end{aligned} $

(5.3.10)

The limit of the sequence defined by the recursive formula for *n* = 0, 1, 2, …, will be found, assuming that *a*
_{0} > 0.

$\displaystyle \begin{aligned} a_{n+1}=\frac{3a_n}{a_n+1}, {} \end{aligned} $

(5.3.11)

In this example, the recursive formula *a*
_{n+1} = *f*(*a*
_{n}) is nonlinear. In this case, we do not have at our disposal any universal method leading to a formula for the general sequence member, but one can still try to find the limit. First of all, it should be noted that if this limit exists (let us denote it with *g* ), one is entitled to execute *n* →*∞* on both sides of the equation (5.3.11). Naturally, the limits of *a*
_{n} and *a*
_{n+1} are identical, and the function on the right-hand side is continuous for arguments that interest us. (If *a*
_{0} > 0, the recurrence will never bring us out of positive values *a*
_{n}, for which the right-hand side of (5.3.11) is well defined.) So one can execute the limit within its argument. In that way, one gets the equation which has two solutions: *g* = 0 and *g* = 2. Therefore, if a limit exists, it must be equal to one of these two numbers.

$\displaystyle \begin{aligned} g=\frac{ 3g}{g+1} , {} \end{aligned} $

(5.3.12)

To investigate the behavior of the sequence in detail, it is convenient to first make a drawing. It not only provides a hint as to the convergence, but also gives us information on how to perform the relevant proof. In Fig. 5.2 the function *y* = *f*(*x*) has been plotted, which in our example is *f*(*x*) = 3*x*∕(*x* + 1), as well as the auxiliary line *y* = *x*. Their intersection has been found above and takes place for arguments *x* = 0 and *x* = 2. For definiteness, suppose that the recursion starts from 0 < *a*
_{0} < 2. If *a*
_{0} = 2, nothing interesting happens because *f*(2) = 2 and the sequence is constant. In turn, the case *a*
_{0} > 2 will be considered later.

Two important observations can easily be made when looking at the picture. If *x* ∈ [0, 2[, then the graph of the function *f* is situated below the straight line *y* = 2 and above *y* = *x*. From the former, one sees that substituting any argument from the interval ]0, 2[ never yields a number greater than 2. Since it will also be positive, as a result of recurrence (5.3.11), one again obtains a value in the range ]0, 2[. It can be then concluded that the sequence is bounded above by the number 2.

From the latter, the inequality *a*
_{n+1} = *f*(*a*
_{n}) > *a*
_{n} is obtained, which means that the sequence is monotonically increasing. In the figure, some gray arrows are drawn, which show how the subsequent terms are generated. Starting from *a*
_{0} on the *x*-axis and moving up, one arrives at *a*
_{1}. Then the procedure is repeated, starting from *a*
_{1}. We can see how the terms are arranged in sequence on the *x*-axis. One can say that they are “attracted” to the point *x* = 2. So, if our figure is drawn correctly, we get the answer: the sequence is increasing and bounded above. It is, therefore, convergent! Below it will be demonstrated in a strict way. The conclusion is that the sequence is convergent and its limit must be the number 2.

- 1.
*We prove the boundedness of the sequence.*This proof is going to be made by induction. It has already been assumed that 0 <*a*_{0}< 2, so it remains to demonstrate the following implication for any index*k*:$\displaystyle \begin{aligned} 0<a_k<2\;\implies\; 0<a_{k+1}<2. {} \end{aligned} $(5.3.13)One has the subsequent equalities:$\displaystyle \begin{aligned} a_{k+1}=f(a_k)= \frac{3a_k}{a_k+1}= 2+\frac{3a_k}{a_k+1}-2=2+\frac{a_k-2}{a_k+1}. {} \end{aligned} $(5.3.14)It is easy to check that under the inductive assumption the expression (

*a*_{k}− 2)∕(*a*_{k}+ 1) is negative. Since*f*(*a*_{k}) > 0, then 0 <*a*_{k+1}< 2, which was to be shown. The sequence is actually bounded. - 2.
*We demonstrate the monotonicity of the sequence.*Now the formula for*a*_{k+1}is going to be transformed as follows:$\displaystyle \begin{aligned} a_{k+1}=f(a_k)= \frac{3a_k}{a_k+1}=\frac{3}{a_k+1}\cdot a_k . {} \end{aligned} $(5.3.15)If 0 <

*a*_{k}< 2, which we already know, the factor 3∕(*a*_{k}+ 1) > 1, and then*a*_{k+1}>*a*_{k}. The sequence is increasing.

Now we are going to consider what changes in our reasoning occur if *a*
_{0} > 2. Let us again refer to the figure: the graph of the function *f* in this interval lies above the line *y* = 2 (i.e., the sequence is bounded below by the number 2) and below the line *y* = *x* (i.e., the sequence is decreasing, because *f*(*a*
_{n}) < *a*
_{n} ). So again, the sequence has to converge and its limit must be the number 2. A strict proof is analogous to that carried out above and one can even use formulas (5.3.14) and (5.3.15). Now, for *a*
_{k} > 2 one has (*a*
_{k} − 2)∕(*a*
_{k} + 1) > 0, and, therefore, also *a*
_{k+1} > 2. Furthermore 3∕(*a*
_{k} + 1) < 1, and in consequence *a*
_{k+1} < *a*
_{k}.

Of course not all functions *f* behave in such a way as shown in the figure. An interesting situation is the case where a sequence is convergent but not monotonic. (This is the case we will look at in the next problem.) In other cases, it may happen that the terms “escape” from the fixed point of a function *f*. Such a sequence is divergent, unless there is another fixed point somewhere else.

The limit of the sequence defined by the recursive formula. for *n* = 0, 1, 2, …, will be found, assuming *a*
_{0} > 0.

$\displaystyle \begin{aligned} a_{n+1}=\frac{6}{2a_n+1}, {} \end{aligned} $

(5.4.1)

As in the previous example, we start this exercise by drawing the appropriate figure. The function defining the recursion will be plotted—in our example it is the function *f*(*x*) = 6∕(2*x* + 1)—as well as the straight line *y* = *x*. It is sufficient to limit oneself to positive values of *x*. This is because if recurrence starts at a certain *a*
_{0} > 0, from the equation (5.4.1), one quickly sees that each subsequent term will be positive too.

The *x* coordinate of the point of intersection of these two graphs (in our case equal to 3∕2 ), i.e., the fixed point of function *f* is a candidate for a limit *g*, which is known from the previous example. If the limit of *a*
_{n} exists, it must be equal to 3∕2.

In the previous problem we were proving that the sequence was monotonic and bounded, and hence one could conclude that it was convergent. In the present example, the effort put into this kind of evidence would be vain—our sequence *is not* monotonic! To realize this, let us refer to Fig. 5.3 where the first few terms are shown. We begin with *a*
_{0} marked on the *x*-axis and move as is indicated by the arrows. Let us chose *a*
_{0} < *g* = 3∕2. When one calculates *f*(*a*
_{0}), it will be seen that it is above the line *y* = 3∕2, and, therefore, *a*
_{1} will be located on the *x*-axis to the right of 3∕2. The next step will lead us again to the left of the fixed point. It is clear that our sequence will oscillate around this point (which is due to the fact that the function *f* is decreasing). Whether the number 3∕2 will be “attracting” the subsequent terms, as is the case in our figure, or “repelling” them as in the next example, depends on a particular function *f*.

Assuming that our figure has been drawn with enough precision, one can see the contents of the convergence proof of *a*
_{n}. It will be composed of the following steps:

- 1.
We separate

*a*_{n}into two subsequences: one with even indexes and one with odd ones. (They are respectively called the even-numbered sequence and the odd-numbered sequence.) - 2.
For each subsequence, the recursive formula with a certain new function $\tilde {f}$ is written down.

- 3.
It is proved that each of these subsequences is monotonic and bounded, and hence convergent.

- 4.
The limits of both subsequences are fixed points of the relevant functions. Both these limits are next verified to be identical.

Since both subsequences “consume” all the terms of *a*
_{n}, their possible common limit will be the limit of the sequence *a*
_{n}.

At this point, one still has to explain the adopted assumption 0 < *a*
_{0} < 3∕2. Well, if one took *a*
_{0} > 3∕2, then in the first step of the recursion one would just get 0 < *a*
_{1} < 3∕2. So all one has to do in this case is to discard the first term with no impact on the limit, and one returns to the sequence considered here. In turn, the value *a*
_{0} = 3∕2 leads to the trivial constant sequence.

Now let us carry out the delineated program. The formula (5.4.1) may be given the form

$\displaystyle \begin{aligned} a_{n+2}=\frac{6}{2a_{n+1}+1}. {} \end{aligned} $

(5.4.2)

Now, again using formula (5.4.1), one can eliminate *a*
_{n+1} for *a*
_{n}, obtaining This recurrence is “by 2” and binds separately even terms and odd terms with one another: The aforementioned function $\tilde {f}$ for both subsequences has, therefore, the form Its only positive fixed point is still *x* = 3∕2. First, the even-numbered subsequence will be considered. The figure suggests that it is necessary to try to prove that this subsequence is bounded and increasing. The conclusion is that the even-numbered subsequence is convergent and its limit must be the number 3∕2.

$\displaystyle \begin{aligned} a_{n+2}=\frac{6}{2\cdot 6/(2a_n+1)+1}=6\;\frac{2a_n+1}{2a_n+13}. {} \end{aligned} $

(5.4.3)

$\displaystyle \begin{aligned} \begin{array}{rcl} a_{2k+2}&\displaystyle =&\displaystyle 6\;\frac{2a_{2k}+1}{2a_{2k}+13} ,{} \end{array} \end{aligned} $

(5.4.4)

$\displaystyle \begin{aligned} \begin{array}{rcl} a_{2k+1}&\displaystyle =&\displaystyle 6\;\frac{2a_{2k-1}+1}{2a_{2k-1}+13}.{} \end{array} \end{aligned} $

(5.4.5)

$\displaystyle \begin{aligned}\tilde{f}(x)=6\;\frac{2x+1}{2x+13}.\end{aligned}$

- 1.
*We prove the boundedness.*As in the previous example, the mathematical induction is going to be used. Given 0 <*a*_{0}< 3∕2, it now ought to be demonstrated that, for any*k*,$\displaystyle \begin{aligned} 0<a_{2k}<\frac{3}{2}\; \implies\; 0<a_{2k+2}<\frac{3}{2}. {} \end{aligned} $(5.4.6)The following series of equalities takes place:$\displaystyle \begin{aligned} \begin{array}{rcl} a_{2k+2}&\displaystyle =&\displaystyle \tilde{f}(a_{2k})= 6\;\frac{2a_k+1}{2a_k+13}=\frac{3}{2}+6\;\frac{2a_k+1}{2a_k+13}-\frac{3}{2}\\ &\displaystyle =&\displaystyle \frac{3}{2}+9\;\frac{a_{2k}-3/2}{2a_{2k}+13}.{} \end{array} \end{aligned} $(5.4.7)It is clear that under the inductive assumption, the second component is negative, and, therefore,

*a*_{2k+2}< 3∕2. Of course*a*_{2k+2}is positive too, which was mentioned at the beginning. It follows that the even-numbered subsequence is in fact bounded. - 2.
*We demonstrate the monotonicity.*Let us transform*a*_{2k+2}as follows:$\displaystyle \begin{aligned} \begin{array}{rcl} a_{2k+2}&\displaystyle =&\displaystyle \tilde{f}(a_{2k})= 6\;\frac{2a_{2k}+1}{2a_{2k}+13} =\frac{12a_{2k}+6}{2a_{2k}+13}\\ &\displaystyle =&\displaystyle \frac{12a_{2k}+6}{2a_{2k}^2+13a_{2k}}\; a_{2k} =\frac{2a_{2k}^2+13a_{2k}-2a_{2k}^2-a_{2k}+6}{2a_{2k}^2+13a_{2k}}\; a_{2k}\\ &\displaystyle =&\displaystyle \frac{2a_{2k}^2+13a_{2k}+2(a_{2k}+2)(3/2-a_{2k})}{2a_{2k}^2+13a_{2k}}\; a_{2k}{}\\ &\displaystyle =&\displaystyle \left(1+2\;\frac{(a_{2k}+2)(3/2-a_{2k})}{2a_{2k}^2+13a_{2k}}\right)a_{2k}. \end{array} \end{aligned} $(5.4.8)For 0 <

*a*_{2k}< 3∕2, which has already been shown, the numerator of the fraction is positive, and, therefore, the entire expression in brackets is larger than 1. Thus,*a*_{2k+2}>*a*_{2k}, i.e., the even-numbered subsequence is increasing.

For the odd-numbered subsequence we prove that it is bounded below by 3∕2 and decreasing. Since the recurrence is described by the identical function $\tilde {f}$, some expressions derived previously can still be used. Rewriting the formula (5.4.7) and adjusting indexes, one has In this case, the inductive assumption has the form *a*
_{2k−1} > 3∕2, which means that the second component is positive. One then gets the inductive thesis: *a*
_{2k+1} > 3∕2. Naturally, in this case the condition *a*
_{1} > 3∕2 required for induction is met, which is due to assumptions 0 < *a*
_{0} < 3∕2. For one has In turn, using the formula (5.4.8), and noting that this time the numerator is negative, one concludes that *a*
_{2k+1} < *a*
_{2k−1}. The odd-numbered subsequence is decreasing and bounded, just what we wanted to show. The limit must constitute the positive fixed point of the function $ \tilde {f} $, which is the number 3∕2.

$\displaystyle \begin{aligned} a_{2k+1}=\frac{3}{2}+9\;\frac{a_{2k-1}-3/2}{2a_{2k-1}+13}. {} \end{aligned} $

(5.4.9)

$\displaystyle \begin{aligned} a_1=\frac{6}{2a_0+1}>\frac{6}{2\cdot 3/2+1}=\frac{3}{2}. {} \end{aligned} $

(5.4.10)

$\displaystyle \begin{aligned} \begin{array}{rcl} a_{2k+1}&\displaystyle =&\displaystyle \frac{2a_{2k-1}^2+13a_{2k-1}+2(a_{2k-1}+2)(3/2-a_{2k-1})}{2a_{2k-1}^2+13a_{2k-1}}\; a_{2k-1}\\ &\displaystyle =&\displaystyle \left(1+2\;\frac{(a_{2k-1}+2)(3/2-a_{2k-1})}{2a_{2k-1}^2+13a_{2k-1}}\right)a_{2k-1},{} \end{array} \end{aligned} $

(5.4.11)

Since both limits are identical, and both subsequences “consume” all terms of *a*
_{n}, then the sequence cannot have other cluster points and one has

$\displaystyle \begin{aligned} \lim_{n\rightarrow\infty}a_n=\frac{3}{2}. {} \end{aligned} $

(5.4.12)

The convergence of the sequence defined by the recursive formula for *n* = 0, 1, 2, …, will be examined. Nothing particular is assumed about $a_0 \in \mathbb {R}$.

$\displaystyle \begin{aligned} a_{n+1}=-a_n^3+2, {} \end{aligned} $

(5.4.13)

From the previous examples, we already know that the eventual limit of *a*
_{n} can only be a fixed point of the function The only such point is *x* = 1, obtained by solving the equation − *x*
^{3} + 2 = *x*. It may be rewritten in the form from which it can be seen that other solutions do not exist because a trinomial in brackets is always positive. (Its discriminant △ = −7 < 0.)

$\displaystyle \begin{aligned} x\;\longmapsto\; f(x)=-x^3+2. {}\end{aligned} $

(5.4.14)

$\displaystyle \begin{aligned} -x^3-x+2=-(x-1)(x^2+x+2)=0 {}\end{aligned} $

(5.4.15)

So if *a*
_{0} = 1, then all *a*
_{n} = 0 and the sequence is constant. This case is not interesting and its solution—obvious. Therefore, we are going to continue with the assumption *a*
_{0} ≠ 1. It should be noted that in this case for all subsequent terms, one still has *a*
_{n} ≠ 1. This is because if one accepted that some *a*
_{n+1} = 1, the following equation would have to be satisfied: and it follows that also *a*
_{n} = 1. By repeating this argumentation now for *a*
_{n} and *a*
_{n−1}, one gets *a*
_{n−1} = 1, and all previous terms, including *a*
_{0}, would have to be equal to unity too.

$\displaystyle \begin{aligned} a_{n+1}=-a_n^3+2=1 , {}\end{aligned} $

(5.4.16)

The situation faced around the fixed point *x* = 1 is shown in Fig. 5.4. One can see that the subsequent members “escape” from this point. Below this conclusion is demonstrated in a strict way, which also means that *a*
_{n} has no limit.

One must remember that if the number *g* is the limit of this sequence, then for any *𝜖* > 0 almost all its members must satisfy the condition |*a*
_{n} − *g*| < *𝜖* (see (5.2.2)). “Almost all” means “all starting from some (generally large) *N*”. Therefore, let us choose some small *𝜖* and suppose that certain *a*
_{n} for large *n* lies in the vicinity of 1. Let us examine what happens to *a*
_{n+1}. Now notice that for *a*
_{n} lying close to 1, and even for any nonnegative *a*
_{n} we have $|a_n^2+a_n+1|\geq 1$, which means that |*a*
_{n+1} − 1|≥|*a*
_{n} − 1|. On the other hand, if (for some *n*) *a*
_{n} were negative, then it would also have to be since all terms on the right-hand side would be negative. Such a sequence—with infinitely many members less than zero—naturally could not be convergent to *g* = 1.

$\displaystyle \begin{aligned} |a_{n+1}-1|=|-a_n^3+2-1|=|a_n^3-1|=|a_n-1|\cdot |a_n^2+a_n+1|. {} \end{aligned} $

(5.4.17)

$\displaystyle \begin{aligned} a_{n+2}=-(\underbrace{-a_n^3+2}_{a_{n+1}})^3+2=a_n^9-6a_n^6+12a_n^3-6<0 , {} \end{aligned} $

(5.4.18)

From the inequality |*a*
_{n+1} − 1|≥|*a*
_{n} − 1|, it follows, therefore, that the sequence cannot have the limit equal to 1. This number “repels” the subsequent terms as is shown in the figure. As discussed previously, other limits of the sequence *a*
_{n} cannot exist. It is then a divergent sequence (for *a*
_{0} ≠ 1).

It will be proved that the sequence is divergent for *n* →*∞*. Its cluster points as well as extremal limits will be examined.

$\displaystyle \begin{aligned} a_n=\cos\left(\frac{\pi n^3}{2n^2+n}\right) {}\end{aligned} $

(5.5.1)

A similar sequence has already been met—it was in Example 4 of Sect. 5.1. We are dealing with a periodic (cosine) expression, the argument of which goes to infinity with *n*. Out of this argument, as we have learned, one ought to extract the leading, divergent term. For large *n*, naturally, one has This quantity should be isolated. It will be achieved by subtracting it and then adding it to the argument of the cosine function. Using the well-known formula $\cos {}(\alpha +\beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta $, one can write For the purpose of simplification, let us write this expression in the form

$\displaystyle \begin{aligned} \frac{\pi n^3}{2n^2+n}\simeq \frac{\pi}{2}\;n. {}\end{aligned} $

(5.5.2)

$\displaystyle \begin{aligned} \begin{array}{rcl} a_n&\displaystyle =&\displaystyle \cos\left(\frac{\pi n^3}{2n^2+n}\right)=\cos\left(\frac{\pi n^3}{2n^2+n}-\frac{\pi n}{2}+\frac{\pi n}{2}\right){}\\ &\displaystyle =&\displaystyle \cos\left(\frac{\pi n^3}{2n^2+n}-\frac{\pi n}{2}\right)\cos\left(\frac{\pi n}{2}\right)-\sin\left(\frac{\pi n^3}{2n^2+n}-\frac{\pi n}{2}\right)\sin\left(\frac{\pi n}{2}\right). \end{array} \end{aligned} $

(5.5.3)

$\displaystyle \begin{aligned} a_n=c_n\cos\left(\frac{\pi n}{2}\right)-s_n\sin\left(\frac{\pi n}{2}\right) , {} \end{aligned} $

(5.5.4)

where two auxiliary sequences have been defined: Now let us transform the new argument of these trigonometric functions as follows: Its infinite limit constitutes the number − *π*∕4, and this—together with the continuity of trigonometric functions—entails It is important that these limits are different from zero. This suggests that the sequence (5.5.4) will not have any limit because the coefficients accompanying *c*
_{n} and *s*
_{n}, i.e., $\cos \left (\pi n/2\right )$ and $\sin \left (\pi n/2 \right ) $, oscillate, taking alternate values: … 0, 1, 0, − 1, 0, 1, 0, − 1, 0, …. In order to demonstrate the divergence of *a*
_{n}, all one must do is to indicate two subsequences convergent to different limits. To start, let us take a subsequence for which *n* will be constantly even (i.e., *n* = 2*m*, $m\in \mathbb {N}$). Then where the known facts that $\sin m\pi =0$, and $\cos m\pi =(-1)^m$ have been used.

$\displaystyle \begin{aligned} \begin{array}{rcl} c_n&\displaystyle :=&\displaystyle \cos\left(\frac{\pi n^3}{2n^2+n}-\frac{\pi n}{2}\right) ,{} \end{array} \end{aligned} $

(5.5.5)

$\displaystyle \begin{aligned} \begin{array}{rcl} s_n&\displaystyle :=&\displaystyle \sin\left(\frac{\pi n^3}{2n^2+n}-\frac{\pi n}{2}\right).{} \end{array} \end{aligned} $

(5.5.6)

$\displaystyle \begin{aligned} \frac{\pi n^3}{2n^2+n}-\frac{\pi n}{2}=\frac{\pi n}{2}\left(\frac{n^2}{n^2+n/2}-1\right)=-\frac{\pi}{4}\cdot \frac{n^2}{n^2+n/2}=-\frac{\pi}{4}\cdot \frac{n}{n+1/2}.\notag\\ {} \end{aligned} $

(5.5.7)

$\displaystyle \begin{aligned} \lim_{n\rightarrow\infty} c_n=\cos\left(-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} ,\;\;\;\; \lim_{n\rightarrow\infty} s_n=\sin\left(-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}. {} \end{aligned} $

(5.5.8)

$\displaystyle \begin{aligned} a_{2m}=c_{2m}\cos m\pi-s_{2m}\sin m\pi=(-1)^mc_{2m} , {}\end{aligned} $

(5.5.9)

The subsequence *a*
_{2m} still has no limit due to the oscillatory factor (−1)^{m}. Our choice was not good enough (after all we wanted to indicate different subsequences that do *converge* to different limits), but we already know how to improve it: instead of considering all *m*’s, one should choose only even, i.e., *n* divisible by 4 ( *n* = 4*k*, $k\in \mathbb {N}$): Since the sequence *c*
_{n} was convergent to $\sqrt {2}/2$, to the same limit must be convergent each of its infinite subsequences (this is a straight conclusion from the definition of the limit), and, therefore, also *c*
_{4k}. Thus, The second subsequence, convergent to a different limit, is easy to find if one comes back to (5.5.9) and select *m* as an odd number, i.e., *n* = 2*m* = 2(2*k* − 1) = 4*k* − 2, for $k\in \mathbb {N}$: In that way, two subsequences convergent to distinct limits have been found: $\pm \sqrt {2}/2$, which ends the proof of the divergence of *a*
_{n}. Comparing (5.5.11) and (5.5.12), one can see why it was so important that the limit of *c*
_{n} was nonzero. For zero value of the limit the oscillations of coefficients would not be relevant and in both cases the limit would be simply ± 0 = 0.

$\displaystyle \begin{aligned} a_{4k}=(-1)^{2k}c_{4k}=c_{4k}. {}\end{aligned} $

(5.5.10)

$\displaystyle \begin{aligned} \lim_{k\rightarrow\infty}a_{4k}=\lim_{k\rightarrow\infty}c_{4k}=\frac{\sqrt{2}}{2}. {} \end{aligned} $

(5.5.11)

$\displaystyle \begin{aligned} a_{4k-2}=(-1)^{2k-1}c_{4k-2}=-c_{4k-2}\underset{k\rightarrow\infty}{\longrightarrow}-\frac{\sqrt{2}}{2}. {} \end{aligned} $

(5.5.12)

Since in this exercise we are interested in extreme limits, let us construct another subsequence, exploiting so far unused members of *a*
_{n}: these are the ones with odd *n* (i.e., *n* = 2*m* − 1, $m\in \mathbb {N}$): As before, there are now two options: to take *m* either even (equal to 2*k*) or odd (equal to 2*k* − 1). One gets It should be noted that each term of *a*
_{n} belongs to one of four subsequences: Therefore, there are no subsequences with limits other than $\pm \sqrt {2}/2$, and consequently, there are no other cluster points. If so, one can write If the set of cluster points (i.e., the set of limits of all convergent subsequences) had more than two elements found in the solution, then to find its extreme limits, we would simply take the least upper and the greatest lower bounds.

$\displaystyle \begin{aligned} a_{2m-1}=c_{2m-1}\cos\left(m-\frac{1}{2}\right)\pi-s_{2m-1}\sin \left(m-\frac{1}{2}\right)\pi=(-1)^ms_{2m-1}. {} \end{aligned} $

(5.5.13)

$\displaystyle \begin{aligned} \begin{array}{rcl} a_{2(2k)-1}&\displaystyle =&\displaystyle a_{4k-1}=(-1)^{2k}s_{4k-1}=s_{4k-1}\underset{k\rightarrow\infty}{\longrightarrow}-\frac{\sqrt{2}}{2} ,{} \end{array} \end{aligned} $

(5.5.14)

$\displaystyle \begin{aligned} \begin{array}{rcl} a_{2(2k+1)-1}&\displaystyle =&\displaystyle a_{4k+1}=(-1)^{2k+1}s_{4k+1}=-s_{4k+1}\underset{k\rightarrow\infty}{\longrightarrow}\frac{\sqrt{2}}{2}.{} \end{array} \end{aligned} $

(5.5.15)

$\displaystyle \begin{aligned} a_{4k-2},\; a_{4k-1},\; a_{4k},\; a_{4k+1} . \end{aligned}$

$\displaystyle \begin{aligned} \limsup_{n\rightarrow\infty} a_n=\frac{\sqrt{2}}{2} ,\;\;\;\; \liminf_{n\rightarrow\infty} a_n= -\frac{\sqrt{2}}{2}.{} \end{aligned} $

(5.5.16)

It will be proved that the sequence is divergent when *n* →*∞*. Its cluster points and extremal limits will also be found.

$\displaystyle \begin{aligned} a_n=\frac{n}{2n+1}\sin\left(\frac{2\pi}{3}\;n\right) {}\end{aligned} $

(5.5.17)

The experience gained in the previous example tells us immediately that the sequence (5.5.17) should be divergent. One sees in fact that *a*
_{n} is in the form of the product of *b*
_{n} = *n*∕(2*n* + 1) whose limit is not zero, but the number 1∕2, and of the oscillating factor $\sin \left (2\pi n/3\right )$. If so, one should be able to indicate at least two subsequences convergent to different limits. What guides us while choosing these subsequences? Well, as in the previous example, one would like to get rid of the oscillating factor because only then the subsequent will be convergent.

Looking at the factor $\sin \left (2\pi n/3\right )$, we conclude that we should start with *n* being a multiple of 3, in order to cancel the denominator. Therefore, let *n* = 3*k* for $k\in \mathbb {N}$: In this way the constant sequence has been obtained, all members of which are equal to zero and consequently

$\displaystyle \begin{aligned} a_{3k}=b_{3k}\sin{}(2\pi k) =0. {}\end{aligned} $

(5.5.18)

$\displaystyle \begin{aligned} \lim_{k\rightarrow\infty}a_{3k}=0. {}\end{aligned} $

(5.5.19)

We are still left to work out indices such as Let us consider, therefore, the subsequence: where, in the successive transformations, the periodicity and odd parity of the sine function have been used as well as the reduction formula:

$\displaystyle \begin{aligned} n=3k-2\;\;\;\; \mathrm{and}\;\;\;\; n=3k-1\;\;\;\;\; \mathrm{for}\;\;k=1,2,3,\ldots {}\end{aligned} $

(5.5.20)

$\displaystyle \begin{aligned} \begin{array}{rcl} a_{3k-2}&\displaystyle =&\displaystyle b_{3k-2}\sin\left(2\pi k-\frac{4\pi}{3}\right)=b_{3k-2}\sin\left(-\frac{4\pi}{3}\right)\\ &\displaystyle =&\displaystyle -b_{3k-2}\,\sin\frac{4\pi}{3}=\frac{\sqrt{3}}{2}\; b_{3k-2}{} , \end{array} \end{aligned} $

(5.5.21)

$\displaystyle \begin{aligned} \sin\frac{4\pi}{3}=-\sin\frac{\pi}{3}=-\frac{\sqrt{3}}{2}. \end{aligned}$

The limit of this subsequence is obvious: Now one should look at the last subsequence: The next limit, different from the previous ones, is obtained:

$\displaystyle \begin{aligned} a_{3k-2}\underset{k\rightarrow\infty}\longrightarrow \frac{\sqrt{3}}{2}\cdot \frac{1}{2}=\frac{\sqrt{3}}{4}\neq 0. {} \end{aligned} $

(5.5.22)

$\displaystyle \begin{aligned} \begin{array}{rcl} a_{3k-1}&\displaystyle =&\displaystyle b_{3k-1}\sin\left(2\pi k-\frac{2\pi}{3}\right)=b_{3k-1}\sin\left(-\frac{2\pi}{3}\right)\\ &\displaystyle =&\displaystyle -b_{3k-1}\,\sin\frac{2\pi}{3}=-\frac{\sqrt{3}}{2}\; b_{3k-1} .{} \end{array} \end{aligned} $

(5.5.23)

$\displaystyle \begin{aligned} a_{3k-1}\underset{k\rightarrow\infty}\longrightarrow \frac{\sqrt{3}}{2}\left(-\frac{1}{2}\right)=-\frac{\sqrt{3}}{4}. {} \end{aligned} $

(5.5.24)

We have then identified three different subsequences convergent to the limits: 0 and $\pm \sqrt {3}/4$. This means that *a*
_{n} is divergent. The subsequences *a*
_{3k−2}, *a*
_{3k−1} and *a*
_{3k} consumed all terms of *a*
_{n}, so the only cluster points are: $\{-\sqrt {3}/4,0,\sqrt {3}/4 \}$, and the extreme limits are as follows:

$\displaystyle \begin{aligned} \limsup_{n\rightarrow\infty} a_n=\frac{\sqrt{3}}{4} ,\;\;\;\; \liminf_{n\rightarrow\infty} a_n= -\frac{\sqrt{3}}{4}.{} \end{aligned} $

(5.5.25)

Exercise 1

Examine the convergence and eventually find limits of sequences:

- (a)
$a_n=\sqrt [3]{n^2+n}-\sqrt [3]{n^2+1}$.

- (b)
$\displaystyle a_n=\frac {\sqrt {n^2+a\,n+1}-\sqrt {n^2+b\,n+2}}{\sqrt {n^2+c\,n+3}-\sqrt {n^2+d\,n+4}}$ , where

*a*,*b*,*c*,*d*> 0.

- (a)
Convergent, limit equal to 0.

- (b)
Convergent, limit equal to (

*a*−*b*)∕(*c*−*d*).

Exercise 2

Using the known criteria, examine the convergence and, eventually, find the limits of sequences:

- (a)
$\displaystyle a_n=\frac {1}{\sqrt {n^4+1}}+\frac {1}{\sqrt {n^4+2}}+\ldots +\frac {1}{\sqrt {n^4+n^2}}$.

- (b)
$\displaystyle a_n=\left (\frac {n+1}{n\sqrt [n]{2}}\right )^{n^2} ,\;\;\;\; b_n=\frac {5^n n!}{(2n)^n}$.

- (c)
$\displaystyle a_n =\sin {}(\pi \sqrt {n^2+2n+2}) ,\;\;\;\; b_n=\cos {}(\pi \sqrt {n^2+2n+2})$.

- (d)
$\displaystyle a_n=\frac {1}{(n+1)!}\, (1\cdot 1!+2\cdot 2!+\ldots +n\cdot n!)$.

- (e)
$\displaystyle a_n=\left (\frac {\sqrt {n+1}+3}{\sqrt {n}+3}\right )^{n^2/(n+1)} ,\;\;\;\; b_n=\left (\cos \frac {\pi }{\sqrt {n}}\right )^n$.

- (a)
Convergent, limit equal to 1.

- (b)
*a*_{n}divergent;*b*_{n}convergent, limit equal to 0. - (c)
*a*_{n}convergent, limit equal to 0;*b*_{n}divergent. - (d)
Convergent, limit equal to 1.

- (e)
*a*_{n}convergent, limit equal to $\sqrt {e}$;*b*_{n}convergent, limit equal to $e^{-\pi ^2/2}$.

Exercise 3

Examine the convergence and eventually find limits of sequences defined recursively:

- (a)
$\displaystyle a_{n+2}=\frac {1}{4}\, (a_{n+1}+3a_n)$, for $\displaystyle a_1=-\frac {3}{2}$ and $\displaystyle a_2=\frac {9}{8}$.

- (b)
$\displaystyle a_{n+1}=\frac {1}{2}\left (a_n+\frac {1}{a_n}\right )$, where

*a*_{1}> 0. - (c)
$\displaystyle a_{n+1}=\frac {8}{2+a_n}$, where

*a*_{1}> 1. - (d)
$\displaystyle a_{n+1}=\frac {1}{a_n^2}$, where

*a*_{1}> 0 and*a*_{1}≠ 1.

- (a)
Convergent, limit equal to 0.

- (b)
Convergent, limit equal to 1.

- (c)
Convergent, limit equal to 2.

- (d)
Divergent.

Exercise 4

Prove that sequences below are divergent:

$\displaystyle \begin{aligned}\displaystyle a_n=\frac{n}{n+1}\,(-1)^{n(n+1)/2} ,\;\;\;\; b_n=(\sqrt{n^2+n}-n)\sin\left(\pi\,\frac{n^2+1}{2n}\right).\end{aligned}$