1.

The following result discussed by Ramanujan is very famous: $$\sqrt[3]{\sqrt[3]{2} - 1} = \sqrt[3]{\frac{1}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{4}{9}}\tag {1}$$ and can be easily proved by cubing both sides and using $x = \sqrt[3]{2}$ for simplified typing.

Ramanujan established many such denesting of radicals such as $$\sqrt{\sqrt[5]{\frac{1}{5}} + \sqrt[5]{\frac{4}{5}}} = \sqrt[5]{1 + \sqrt[5]{2} + \sqrt[5]{8}} = \sqrt[5]{\frac{16}{125}} + \sqrt[5]{\frac{8}{125}} + \sqrt[5]{\frac{2}{125}} - \sqrt[5]{\frac{1}{125}}\tag {2}$$$$\sqrt[3]{\sqrt[5]{\frac{32}{5}} - \sqrt[5]{\frac{27}{5}}} = \sqrt[5]{\frac{1}{25}} + \sqrt[5]{\frac{3}{25}} - \sqrt[5]{\frac{9}{25}}\tag {3}$$$$\sqrt[4]{\frac{3 + 2\sqrt[4]{5}}{3 - 2\sqrt[4]{5}}} = \frac{\sqrt[4]{5} + 1}{\sqrt[4]{5} - 1}\tag{4}$$$$\sqrt[\color{red}6]{7\sqrt[3]{20} - 19} = \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}}\tag{5}$$$$\sqrt[6]{4\sqrt[3]{\frac{2}{3}} - 5\sqrt[3]{\frac{1}{3}}} = \sqrt[3]{\frac{4}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{1}{9}}\tag{6}$$

$$\sqrt[8]{1+\sqrt{1-\left(\frac{-1+\sqrt{5}}{2}\right)^{24}}} = \frac{-1+\sqrt{5}}{2}\,\frac{1+\sqrt[4]{5}}{\sqrt{2}}\tag{7}$$

with the last one found in *Ramanujan's Notebooks*, Vol 5, p. 300. Most of these radical expressions are **units** (a unit is an algebraic integer $\alpha$ such that $\alpha\beta = 1$ where $\beta$ is another algebraic integer).

For me the only way to establish these identities is to raise each side of the equation to an appropriate power using brute force algebra and then check the equality. However for higher powers (for example equation $(2)$ above) this seems very difficult.

> Is there any underlying structure in these powers of units which gives rise to such identities or these are mere strange cases which were noticed by Ramanujan who used to play with all sorts of numbers as a sort of hobby? I believe (though not certain) that perhaps Ramanujan did have some idea of such structure which leads to some really nice relationships between units and their powers. I wonder if there is any sound theory of such relationships which can be exploited to give many such identities between nested and denested radicals.

We also have the following identity,

$$\sqrt[3]{m^3-n^3+6m^2n+3mn^2-3(m^2+mn+n^2)\sqrt[3]{mn(m+n)}}=\\ \sqrt[3]{m^2(m+n)}-\sqrt[3]{mn^2}-\sqrt[3]{(m+n)^2n}$$

For $m=n=1$ we get $(1)$.

For $m=4$ and $n=1$ we get $(5)$.

Wikipedia [informs][1] me that

$$S = \vartheta(0;i)=\sum_{n=-\infty}^\infty e^{-\pi n^2} = \frac{\sqrt[4] \pi}{\Gamma\left(\frac 3 4\right)}$$

I tried considering $f(x,n) = e^{-x n^2}$ so that its Mellin transform becomes $\mathcal{M}_x(f)=n^{-2z} \Gamma(z)$ so inverting and summing

$$\frac{1}{2}(S-1)=\sum_{n=1}^\infty f(\pi,n)=\sum_{n=1}^\infty \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}n^{-2z} \Gamma(z)\pi^{-z}\,dz = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\zeta(2z) \Gamma(z) \pi^{-z}\,dz$$

However, this last integral (whose integrand has poles at $z=0,\frac{1}{2}$ with respective residues of $-\frac 1 2$ and $\frac 1 2$) is hard to evaluate due to the behavior of the function as $\Re(z)\to \pm\infty$ which makes a classic infinite contour over the entire left/right plane impossible.

How does one go about evaluating this sum?

[1]: http://en.wikipedia.org/wiki/Theta_function#Explicit_values

This one is a direct evaluation of elliptic integrals. Jacobi's theta function $\vartheta_{3}(q)$ is defined via the equation $$\vartheta_{3}(q) = \sum_{n = -\infty}^{\infty}q^{n^{2}}\tag{1}$$ Let $0 < k < 1$ and $k' = \sqrt{1 - k^{2}}$ and we define elliptic integrals $K, K'$ via $$K(k) = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - k^{2}\sin^{2}x}}, K = K(k), K' = K(k')\tag{2}$$ Then it is [almost a miracle][1] that we can get $k$ in terms of $K, K'$ via the variable $q = e^{-\pi K'/K}$ using equations $$k = \frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)}\tag{3}$$ where $\vartheta_{2}(q)$ is another theta function of Jacobi defined by $$\vartheta_{2}(q) = \sum_{n = -\infty}^{\infty}q^{(n + (1/2))^{2}}\tag{4}$$ Also the function $\vartheta_{3}(q)$ is directly related to $K$ via $$\vartheta_{3}(q) = \sqrt{\frac{2K}{\pi}}\tag{5}$$ The proofs of $(3)$ and $(5)$ are given in the linked post on my blog.

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The sum in the question is $\vartheta_{3}(e^{-\pi})$ so that we have $q = e^{-\pi}$. This implies that $K'/K = 1$ so that $k = k'$ and from $k^{2} + k'^{2} = 1$ we get $k^{2} = 1/2$. And then $$\vartheta_{3}(q) = \sqrt{\frac{2K}{\pi}} = \sqrt{\frac{2}{\pi}\cdot\frac{\Gamma^{2}(1/4)}{4\sqrt{\pi}}} = \frac{\Gamma(1/4)}{\pi^{3/4}\sqrt{2}}$$ Now using $\Gamma(1/4)\Gamma(3/4) = \pi/\sin(\pi/4) = \pi\sqrt{2}$ we get $$\sum_{n = -\infty}^{\infty}e^{-\pi n^{2}} = \vartheta_{3}(e^{-\pi}) = \frac{\sqrt[4]{\pi}}{\Gamma(3/4)}$$ The value of $K = K(1/\sqrt{2})$ in terms of $\Gamma(1/4)$ is evaluated in [this answer][2].

[1]: http://paramanands.blogspot.com/ ... heta-functions.html

[2]: https://math.stackexchange.com/a/1793756/72031

The integral $\displaystyle\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}}$ is equal to $\displaystyle \frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi }}$.

It is calculated or verified with a computer algebra system that $\displaystyle \frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi }} = K\left(\frac{1}{2}\right)$ , where $K(m)$ is the complete elliptic integral of the first kind. This is in relation to what is called the [elliptic integral singular value][1].

It is also known or verified that

$\displaystyle K\left(\frac{1}{2}\right) =\displaystyle \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{1-\frac{\sin ^2(t)}{2}}} \, dt= \frac{1}{2} \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{\sin (t) \cos (t)}} \, dt$.

Can one prove directly or analytically that

$\displaystyle\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}} =\frac{1}{2} \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{\sin (t) \cos (t)}} \, dt =\displaystyle \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{1-\frac{\sin ^2(t)}{2}}} \, dt = K\left(\frac{1}{2}\right) $ ?

[1]: http://mathworld.wolfram.com/EllipticIntegralSingularValue.html

We have

$$\int_{0}^{+\infty}\frac{dx}{\sqrt{(x^2+a^2)(x^2+b^2)}}=\frac{\pi}{2\,\text{AGM}(a,b)}=\frac{\pi}{2\,\text{AGM}\left(\frac{a+b}{2},\sqrt{ab}\right)}\tag{1} $$

[as a consequence][1] of Lagrange's identity $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$.<br>

On the other hand

$$ K\left(\tfrac{1}{2}\right)=\int_{0}^{+\infty}\frac{dx}{\sqrt{(1+x^2)(1+x^2/2)}} = \frac{\pi}{2\,\text{AGM}\left(1,\frac{1}{\sqrt{2}}\right)}\tag{2} $$

and $$\text{AGM}\left(\sqrt{i},\sqrt{-i}\right)=\text{AGM}\left(1,\tfrac{1}{\sqrt{2}}\right) \tag{3}$$

so

$$K\left(\tfrac{1}{2}\right)=\int_{0}^{+\infty}\frac{dx}{\sqrt{(1+x^2)(1+x^2/2)}} =\int_{0}^{+\infty}\frac{dx}{\sqrt{(x^2+i)(x^2-i)}} = \int_{0}^{+\infty}\frac{dx}{\sqrt{x^4+1}}$$

as claimed. You can find an alternative, very short proof through Fourier-Legendre series expansions at page $29$ [here](https://arxiv.org/abs/1710.03221).

[1]: https://math.stackexchange.com/q ... ies/1904382#1904382

Arithmetic Geometric Mean can be represented by a Hypergeometric function:

$$\text{agm}(1,p)=\frac{1}{{_2F_1} \left(\frac{1}{2},\frac{1}{2};1;1-p^2 \right)}$$

$$0<p \leq 1$$

One of the main properties of the AGM is the following identity:

$$\text{agm}(1,p)=\frac{1+p}{2}\text{agm} \left(1,\frac{2\sqrt{p}}{1+p} \right)$$

This allows the infinite product representation of the AGM.

> I wanted to know if it's possible to prove this identity by directly using the Hypergeometric series.

For the Hypergeometric function the identity will take the following form:

$${_2F_1} \left(\frac{1}{2},\frac{1}{2};1;1-p^2 \right)=\frac{2}{1+p} {_2F_1} \left(\frac{1}{2},\frac{1}{2};1;\frac{(1-p)^2}{(1+p)^2} \right)$$

In the series form it will be:

$$ \sum_{k=0}^\infty \frac{1}{k!^2} \left(\frac{1}{2}\right)^2_k (1-p^2)^k=\frac{2}{1+p} \sum_{k=0}^\infty \frac{1}{k!^2} \left(\frac{1}{2}\right)^2_k \frac{(1-p)^{2k}}{(1+p)^{2k}}$$

I think the following substitution will simplify things:

$$1-p=2x$$

> $$ \sum_{k=0}^\infty \frac{1}{k!^2} \left(\frac{1}{2}\right)^2_k 2^{2k}x^k(1-x)^k=\frac{1}{1-x} \sum_{k=0}^\infty \frac{1}{k!^2} {\left(\frac{1}{2}\right)^2_k} \frac{x^{2k}}{(1-x)^{2k}}$$

I haven't been able to prove this identity from the series.

Comparing terms in this form is useless, since the partial sums of the series are not equal (the second series converges much faster).

The only idea I have is to use the uniqueness of the power series, which requires expanding everything, so there are only powers of $x$ left.

We have:

$$(1-x)^k=\sum_{l=0}^k (-1)^l \left(\begin{matrix} k \\ l \end{matrix} \right) x^l$$

$$\frac{1}{(1-x)^{2k+1}}=(2k)! \sum_{n=0}^\infty (-1)^{n+2k}~ (n+2k)_{2k} ~x^n$$

Here $(n+2k)_{2k}$ actually means falling factorial, not rising factorial, like above. $(n+2k)_{2k}=(n+2k)(n+2k-1)(n+2k-2) \cdots$. I don't know what other notation to use in this case.

Now I'm stuck. I don't know how to get the single power series for $x$ on each side so we can compare them.

Use the function $$f(x) = \frac{1}{\operatorname{agm}(1 + x, 1 - x)}\tag{1}$$ which has the property that $$f(x) = \frac{1}{1 + x}\cdot f\left(\frac{2\sqrt{x}}{1 + x}\right)\tag{2}$$ and use the series expansion $$f(x) = 1 + a_{1}x^{2} + a_{2}x^{4} + \cdots\tag{3}$$ (note that the function $f$ is even) and find coefficients $a_{n}$ by using series expansion $(3)$ in functional equation $(2)$. This is how Gauss derived the formula $$f(x) = {}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2};1; x^{2}\right)\tag{4}$$ See [this post][1] of mine for details. Note also that $$f(x) = \frac{1}{\operatorname{agm}(1, \sqrt{1 - x^{2}})}\tag{5}$$ and hence on putting $1 - x^{2} = p^{2}$ we get $$\operatorname{agm}(1, p) = \dfrac{1}{{}_{2}F_{1}\left(\dfrac{1}{2}, \dfrac{1}{2};1; 1 - p^{2}\right)}\tag{6}$$

BTW the evaluation of coefficients $a_{n}$ for general $n$ is difficult but Gauss solved it completely. See [this paper][2] for more details of the calculation of $a_{n}$.

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Another route to the relation between hypergeometric series is to use the differential equation satisfied by the function $y(x) = {}_{2}F_{1}(a, b; c; x)$. The differential equation satisfied by $y$ is given as $$x(1 - x)y'' + \{c - (a + b + 1)x\}y' - aby = 0\tag{7}$$ Let's just write $F$ in place of ${}_{2}F_{1}$ and then we see that $y = F(a, b; 2b; x)$ satisfies the equation $$x(1 - x)y'' + \{2b - (a + b + 1)x\}y' - aby = 0$$ Putting $x = 4z/(1 + z)^{2}$ we can see that that $$\frac{dF}{dx} = \frac{(1 + z)^{3}}{4(1 - z)}\cdot\frac{dF}{dz}$$ and $$\frac{d^{2}F}{dx^{2}} = \frac{(1 + z)^{5}}{16(1 - z)^{3}}\left((1 - z^{2})\frac{d^{2}F}{dz^{2}} + (4 - 2z)\frac{dF}{dz}\right)$$ and $$x(1 - x) = \frac{4z(1 - z)^{2}}{(1 + z)^{4}}$$ and thus after some algebraic manipulation we get $$z(1 - z)(1 + z)^{2}\frac{d^{2}F}{dz^{2}} + 2(1 + z)(b - 2az + bz^{2} - z^{2})\frac{dF}{dz} - 4ab(1 - z)F = 0$$ and the function $F(a, b; 2b; 4z/(1 + z)^{2})$ satisfies this equation. If we put $F = (1 + z)^{2a}G$ we get $$z(1 - z^{2})\frac{d^{2}G}{dz^{2}} + 2\{b - (2a - b + 1)z^{2}\}\frac{dG}{dz} - 2az(1 + 2a - 2b)G = 0$$ It is easily seen that $G(-z)$ also satisfies this equation and hence $G$ is an even function and we can put $z^{2} = t$ to get $$t(1 - t)\frac{d^{2}G}{dt^{2}} + \left(b + \frac{1}{2} - \left(2a - b + \frac{3}{2}\right)t\right)\frac{dG}{dt} - a\left(a - b + \frac{1}{2}\right)G = 0\tag{8}$$ Comparing this with equation $(7)$ we see that solution $G$ is given by $$G = F(a, a - b + 1/2; b + 1/2; t) = F(a, a - b + 1/2; b + 1/2; z^{2})\tag{9}$$ However the way we reached equation $(8)$ shows us that its solution is given by $$G = (1 + z)^{-2a}F(a, b; 2b; 4z/(1 + z)^{2})\tag{10}$$ Both the solutions $(9)$ and $(10)$ are analytic in neighborhood of $0$ and they are equal to $1$ at $z = 0$ and therefore they are equal and we get the quadratic transformation of Gauss $$F\left(a, b; 2b; \frac{4z}{(1 + z)^{2}}\right) = (1 + z)^{2a}F\left(a, a - b + \frac{1}{2}; b + \frac{1}{2}; z^{2}\right)\tag{11}$$ Putting $a = 1/2, b = 1/2$ we get $$F\left(\frac{1}{2}, \frac{1}{2}; 1; \frac{4z}{(1 + z)^{2}}\right) = (1 + z)F\left(\frac{1}{2}, \frac{1}{2}; 1; z^{2}\right)\tag{12}$$ and putting $z = (1 - p)/(1 + p)$ so that $p = (1 - z)/(1 + z)$ we get $$F\left(\frac{1}{2}, \frac{1}{2}; 1; 1 - p^{2}\right) = \frac{2}{1 + p}F\left(\frac{1}{2}, \frac{1}{2}; 1; \frac{(1 - p)^{2}}{(1 + p)^{2}}\right)\tag{13}$$ which is the result in question. Using similar technique we can prove another quadratic transformation $$F\left(a, b; a + b + \frac{1}{2}; 4z(1 - z)\right) = F\left(2a, 2b; a + b + \frac{1}{2}; z\right)\tag{14}$$ See [this post][3] and [the next one][4] for more details.

[1]: http://paramanands.blogspot.com/ ... -mean-of-gauss.html

[2]: http://dl.dropbox.com/u/143641614/pdfs/gaussagm.pdf

[3]: http://paramanands.blogspot.com/ ... etric-series-1.html

[4]: http://paramanands.blogspot.com/ ... etric-series-2.html

I think it is faster to go through an elliptic integral. Given $a,b\in\mathbb{R}^+$ we may define

$$ E(a,b)=\int_{0}^{+\infty}\frac{dx}{\sqrt{(a^2+x^2)(b^2+x^2)}} \tag{1} $$

and notice that by Lagrange's identity $(a^2+x^2)(b^2+x^2)=(ax+bx)^2+(ab-x^2)^2$ and a suitable change of variable we have

$$ E(a,b)=E\left(\frac{a+b}{2},\sqrt{ab}\right) \tag{2}$$

hence:

$$ E(a,b)=E(AGM(a,b),AGM(a,b))=\frac{1}{AGM(a,b)}\int_{0}^{+\infty}\frac{dx}{x^2+1}\tag{3} $$

and

$$ AGM(a,b) = \frac{\pi}{2 E(a,b)}\tag{4} $$

so our claim is equivalent to proving that

$$\frac{2}{\pi}\int_{0}^{+\infty}\frac{dx}{\sqrt{(1+x^2)(p^2+x^2)}} = \phantom{}_2 F_1\left(\frac{1}{2},\frac{1}{2};1;1-p^2\right) \tag{5} $$

but by setting $u=\frac{1}{1+x^2}$, that simply follows from [Euler's integral representation][1] for $\phantom{}_2 F_1$.

[1]: https://en.wikipedia.org/wiki/Hypergeometric_function#Euler_type