5. Investigating Convergence of Sequences and Looking for Their Limits
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 an, starting from the first one, i.e., a1.
If only inequalities an+1 ≥ an or an+1 ≤ an hold, the sequences are called nondecreasing or nonincreasing respectively. Sequences with either of these properties are called monotonic.
The sequence is bounded above if
$\displaystyle \begin{aligned} \exists_{M\in\mathbb{R}}\forall_{n\in \mathbb{N}}\;\; a_{n}\leq M, {} \end{aligned} $
(5.0.3)
and bounded below if
$\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 bn = a2n denotes the subsequence created by deleting all odd-numbered terms of the sequence an.
The sequence can be specified by an explicit formula for an, 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 a1 and provide the dependence an+1 = f(an). The more complicated dependence, for instance an+2 = f(an+1, an), requires fixing two first elements, a1 and a2, etc.
The principal notion for the infinite sequence is its limit. A sequence is called convergent to a certain number g if
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.
For convergent sequences, the following equations hold:
limn→∞an ±limn→∞bn =limn→∞(an ± bn),
(limn→∞an) ⋅ (limn→∞bn) =limn→∞(an ⋅ bn),
(limn→∞an)∕(limn→∞bn) =limn→∞(an∕bn), if limn→∞bn ≠ 0,
and inserting into the second equation a constant sequence for bn, i.e., bn = c, one also gets the relation limn→∞can = climn→∞an.
It can also be proved that a sequence in the special form
A number a is called the cluster point of a sequence an 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.
5.1 Some Common Tricks Useful for Calculating Limits of Sequences
The sequence an 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 an 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 an defined by (5.1.1).
Let us look at the first term. Under the fourth root the polynomial n4 + 2n3 is found. For very large n the leading term is n4, 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 an.
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) = a2 − b2. Let our a and b be, respectively, the first and the second terms in the formula for an. 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!
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:
$\displaystyle \lim _{n\rightarrow \infty }\sqrt [k]{1+a/n}=\sqrt [k]{\lim _{n\rightarrow \infty }(1+a/n)}=\sqrt [k]{1}=1$, 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 cn (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 n4. 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:
As usual, we are going to start solving the problem by carefully looking at the formula for an. 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 an 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 an 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.
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 bn and cn are found with a common limit g, satisfying the inequality bn ≤ an ≤ cn for almost all n, then limn→∞an = g.
Thus, it remains now to choose sequences bn and cn. How can one guess them? Well, they must converge to the common limit, which will also be the limit of the tested sequence an. 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 an, 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 limn→∞an =limn→∞n∕n = 1 = g. It remains to strictly demonstrate this.
Now one has to look for bn and cn. 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 cn:
In this way, it is clear that the assumptions of the squeeze theorem are satisfied. The conclusion is then the limn→∞an = 1.
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 an 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
where $k\in \mathbb {N}$, and λi > 0, i = 1, 2, …, k, will be proved and its limit will be found.
Solution
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 λi’s are not all different for various i, but this has no effect on our reasoning. Let us denote
The exponentials of the type λn 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 λi’s. Therefore, we suspect that
Assuming that the value of the limit has been guessed correctly, one must now indicate two sequences bn and cn, still convergent to λmax and satisfying—for almost all n—the inequalities bn ≤ an ≤ cn.
The choice of bn is clear. Since it is anticipated that the limit of an is determined by the greatest number of λi’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 bn < an holds. In turn, in order to select cn, 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 λi’s equal to λmax:
Of course an < cn and the squeeze theorem gives the expected result: limn→nan = λmax.
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 cn without changes, and choosing bn as
If not all λi are equal, then bn and cn 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.
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 an’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, 2π[ by subtracting from it the appropriate multiple of the period equal to 2π. 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 an, 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
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 an 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.
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:
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 bn means a bounded expression (by a number M > 0) and cn converges to zero, one has
has proved very important. If the sequence cn converged to some g ≠ 0, two subsequences of an could be created: one composed of terms with even indexes (a2k) and the other with those of odd indexes (a2k+1), of which the former goes to g, and the latter to − g. This would mean that the sequence an 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
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.
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:
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 ϕn convergent to zero. In particular, one can choose ϕn = an∕n obtaining (5.1.35).
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.
Fig. 5.1
Comparison of values ϕ, $ \sin \phi $, and $ \tan \phi $
It shows a circle of radius 1 and two rectangular triangles: AOB and COD with the common angle ϕ. From △ AOB one has
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:
will be examined, depending on the values of the parameters $l,k\in \mathbb {N}$, where pk(x) = α0 + α1x + … + αkxk and ql(x) = β0 + β1x + … + βlxl are real polynomials. It is assumed that at least αk and βl do not vanish. In the case of convergence, the limit of an will be found.
Solution
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:
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
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 nk∕nl = 1∕nl−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
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., βl. 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., |βl|∕2. Then it could be written that the denominator is estimated from below as follows:
Because the number of terms of the form βi∕nl−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 |βl|∕2.
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 βl ≠ 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
With this choice, (5.2.9) is actually fulfilled and the proof is complete.
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):
we obtain |an| > M. Almost all terms of the sequence |an| are greater than the arbitrary constant! This sequence cannot, therefore, be bounded. It must be divergent, and the same must refer to the sequence an itself.
Case 3: l = k
Our purpose will be to demonstrate that the limit constitutes the number g = αk∕βk. Therefore, let us consider the difference an − 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.
Problem 2
Applying Cauchy’s test, the convergence of the sequence
In accordance with the text of the problem, we are going to solve it using Cauchy’s criterion called also the “nth 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
$\displaystyle \lim _{n\rightarrow \infty }\sqrt [n]{|a_n|}<1$, then limn→∞an = 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.
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 nth root appears and
$\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)
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.
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
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 3n, so for sufficiently large n one can estimate
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.
Problem 3
Using d’Alembert’s criterion, the convergence of the sequence
$\displaystyle \lim _{n\rightarrow \infty }\left |\frac {a_{n+1}}{a_n}\right |>1$, the sequence an 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.
As it is easy to understand, this criterion is particularly convenient if the formula for an contains problematic factors that cancel or lead to the known limits when dividing an+1 by an. 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
and hence, by virtue of the d’Alembert criterion, it can be deduced that limn→∞an = 0.
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 |an| behaves “worse” than a geometric sequence of quotient M. |an| (and, therefore, also an) 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 an can be convergent (even for q = ∞), which can be found out while considering the example of bn with the following consecutive terms:
when both bn and cn go to infinity, and the sequence cn 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 $.
There appears a quite natural question: why should it be more comfortable studying the behavior of the fraction (bn+1 − bn)∕(cn+1 − cn) than the original expression bn∕cn? Well, there are many examples of sequences that contain sums of many terms in the numerator or denominator. Creating a difference of the type bn+1 − bn, 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,
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
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 an is of the special form
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., n3). 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:
First, let us concentrate on bn. In the numerator we recognize that in addition to n3, 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 n3 ± M, where M is a constant. For the limit of an, this constant is completely irrelevant, and hence one could write in the numerator n3 ± 1 or even simply n3. What could eventually prove to be important is a term behaving as n2 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 $n^3\left (1 +1/n\right )$.
Now the question may arise: how do we know that terms such as n2 (or worse) appearing with n3 are important, while others can be omitted with impunity? Well, the answer is given by the form of the sequence cn together with the condition (5.2.42). For large n one has cn ≃ n, and a finite limit g in (5.2.42) would be obtained when bn ≃ 1∕n. Since both leading expressions in the numerator and the denominator are of the form n3, these giving (at the end) finite limit are n3 ⋅ 1∕n = n2. 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 an, we are going to look for the limit of
Let us now return to the full formula for an 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
for n = 0, 1, 2, …, will be calculated, where the two initial members are chosen as a0 = 5 and a1 = −1∕2.
Solution
The sequence (5.3.1) is defined by a linear recursive equation, which is homogeneous (i.e., without free terms), and the factors multiplying ak 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 an = λn, 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 an 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: λ1 = 1∕2 and λ2 = −1∕4. “Forgetting” for a moment that one has to comply with initial conditions, there are two possibilities:
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 an 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 an for n →∞ is equal to 0.
What complications can occur while solving this type of problem?
The equation for a constantλ, similar to (5.3.2), may have a double rootλ = λ0.
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 λ0 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 formula
The equation for a constantλmay 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 a0 and a1 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 constantλmay 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.
At the end, it is worth sharing where the idea of postulating solutions in the form of λn originates. Well, this concept is taken from matrix calculations. To find that out, let us create the following vector vn from the members of an:
Moving to vn+1 reduces to multiplying the vector vn by a matrix of constant numbers. One can, therefore, immediately write the general formula for vn: vn = Mnv0, where v0 is a vector of initial conditions
and two eigenvalues: λ1 = 1∕2 and λ2 = −1∕4. Consequently, the equation (5.3.2) turns out to be simply a characteristic equation for the matrix M, and parameters λ1,2—its eigenvalues!
If one denotes the eigenvectors with symbols u1 and u2, the subsequent steps of the solution are already clear from the course of algebra: v0 is expanded in the basis of eigenvectors v0 = c1u1 + c2u2 (constants c1,2 are then known) and it can be written as
for n = 0, 1, 2, …, will be found, assuming that a0 > 0.
Solution
In this example, the recursive formula an+1 = f(an) 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 an and an+1 are identical, and the function on the right-hand side is continuous for arguments that interest us. (If a0 > 0, the recurrence will never bring us out of positive values an, 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.
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) = 3x∕(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 < a0 < 2. If a0 = 2, nothing interesting happens because f(2) = 2 and the sequence is constant. In turn, the case a0 > 2 will be considered later.
Fig. 5.2
Generation of subsequent terms in the recurrence (5.3.11)
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 an+1 = f(an) > an 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 a0 on the x-axis and moving up, one arrives at a1. Then the procedure is repeated, starting from a1. 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.
1.
We prove the boundedness of the sequence.
This proof is going to be made by induction. It has already been assumed that 0 < a0 < 2, so it remains to demonstrate the following implication for any index k:
It is easy to check that under the inductive assumption the expression (ak − 2)∕(ak + 1) is negative. Since f(ak) > 0, then 0 < ak+1 < 2, which was to be shown. The sequence is actually bounded.
2.
We demonstrate the monotonicity of the sequence.
Now the formula for ak+1 is going to be transformed as follows:
If 0 < ak < 2, which we already know, the factor 3∕(ak + 1) > 1, and then ak+1 > ak. The sequence is increasing.
The conclusion is that the sequence is convergent and its limit must be the number 2.
Now we are going to consider what changes in our reasoning occur if a0 > 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(an) < an ). 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 ak > 2 one has (ak − 2)∕(ak + 1) > 0, and, therefore, also ak+1 > 2. Furthermore 3∕(ak + 1) < 1, and in consequence ak+1 < ak.
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.
5.4 When a Sequence Oscillates
Problem 1
The limit of the sequence defined by the recursive formula.
for n = 0, 1, 2, …, will be found, assuming a0 > 0.
Solution
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∕(2x + 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 a0 > 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 an 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 a0 marked on the x-axis and move as is indicated by the arrows. Let us chose a0 < g = 3∕2. When one calculates f(a0), it will be seen that it is above the line y = 3∕2, and, therefore, a1 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.
Fig. 5.3
Generation of subsequent terms in the recurrence (5.4.1)
Assuming that our figure has been drawn with enough precision, one can see the contents of the convergence proof of an. It will be composed of the following steps:
1.
We separate an 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 an, their possible common limit will be the limit of the sequence an.
At this point, one still has to explain the adopted assumption 0 < a0 < 3∕2. Well, if one took a0 > 3∕2, then in the first step of the recursion one would just get 0 < a1 < 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 a0 = 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
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.
1.
We prove the boundedness.
As in the previous example, the mathematical induction is going to be used. Given 0 < a0 < 3∕2, it now ought to be demonstrated that, for any k,
It is clear that under the inductive assumption, the second component is negative, and, therefore, a2k+2 < 3∕2. Of course a2k+2 is positive too, which was mentioned at the beginning. It follows that the even-numbered subsequence is in fact bounded.
For 0 < a2k < 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, a2k+2 > a2k, i.e., the even-numbered subsequence is increasing.
The conclusion is that the even-numbered subsequence is convergent and its limit must be the number 3∕2.
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 a2k−1 > 3∕2, which means that the second component is positive. One then gets the inductive thesis: a2k+1 > 3∕2. Naturally, in this case the condition a1 > 3∕2 required for induction is met, which is due to assumptions 0 < a0 < 3∕2. For one has
and noting that this time the numerator is negative, one concludes that a2k+1 < a2k−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.
Since both limits are identical, and both subsequences “consume” all terms of an, then the sequence cannot have other cluster points and one has
from which it can be seen that other solutions do not exist because a trinomial in brackets is always positive. (Its discriminant △ = −7 < 0.)
So if a0 = 1, then all an = 0 and the sequence is constant. This case is not interesting and its solution—obvious. Therefore, we are going to continue with the assumption a0 ≠ 1. It should be noted that in this case for all subsequent terms, one still has an ≠ 1. This is because if one accepted that some an+1 = 1, the following equation would have to be satisfied:
and it follows that also an = 1. By repeating this argumentation now for an and an−1, one gets an−1 = 1, and all previous terms, including a0, would have to be equal to unity too.
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 an has no limit.
Fig. 5.4
The subsequent terms of the divergent sequence defined by (5.4.13)
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 |an − 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 an for large n lies in the vicinity of 1. Let us examine what happens to an+1.
Now notice that for an lying close to 1, and even for any nonnegative an we have $|a_n^2+a_n+1|\geq 1$, which means that |an+1 − 1|≥|an − 1|. On the other hand, if (for some n) an 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.
From the inequality |an+1 − 1|≥|an − 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 an cannot exist. It is then a divergent sequence (for a0 ≠ 1).
is divergent for n →∞. Its cluster points as well as extremal limits will be examined.
Solution
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
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 cn and sn, 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 an, 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 = 2m, $m\in \mathbb {N}$). Then
where the known facts that $\sin m\pi =0$, and $\cos m\pi =(-1)^m$ have been used.
The subsequence a2m 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 = 4k, $k\in \mathbb {N}$):
Since the sequence cn 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 c4k. 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 = 2m = 2(2k − 1) = 4k − 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 an. Comparing (5.5.11) and (5.5.12), one can see why it was so important that the limit of cn 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.
Since in this exercise we are interested in extreme limits, let us construct another subsequence, exploiting so far unused members of an: these are the ones with odd n (i.e., n = 2m − 1, $m\in \mathbb {N}$):
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.
is divergent when n →∞. Its cluster points and extremal limits will also be found.
Solution
The experience gained in the previous example tells us immediately that the sequence (5.5.17) should be divergent. One sees in fact that an is in the form of the product of bn = n∕(2n + 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 = 3k for $k\in \mathbb {N}$:
We have then identified three different subsequences convergent to the limits: 0 and $\pm \sqrt {3}/4$. This means that an is divergent. The subsequences a3k−2, a3k−1 and a3k consumed all terms of an, so the only cluster points are: $\{-\sqrt {3}/4,0,\sqrt {3}/4 \}$, and the extreme limits are as follows:
