I recently saw this question ... https://math.stackexchange.com/questions/2393668/double-factorial-sum-k-0-infty-left-frac-2k-1-2k-right#2393668 ... & I am unable to show it.

The result $$1-\left( \frac{1}{2} \right)^{3}+\left( \frac{1\cdot 3}{2\cdot 4} \right)^{3}-\left( \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6} \right)^{3}+\left( \frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8} \right)^{3}-...=\frac{\Gamma ^{2}\left( \frac{9}{8} \right)}{\Gamma ^{2}\left( \frac{7}{8} \right)\cdot \Gamma ^{2}\left( \frac{10}{8} \right)}$$

Now I know several results relating to those ratios of double factorials \begin{eqnarray*} \int_0^{\frac{\pi}{2}} \sin^{2n} \theta d \theta= \frac{ \pi}{2} \frac{(2n-1)!!}{(2n)!!} \tag{1} \\ \sum_{n=0}^{\infty} \frac{(2n-1)!!}{(2n)!!} y^n = \frac{1}{\sqrt{1-y}} \tag{2} \end{eqnarray*} From these two results it is reasonably easy to derive the series for the elliptic integral of the first kind \begin{eqnarray*} K(k)=\int_0^{\frac{\pi}{2}} \frac{d \theta}{\sqrt{1-k^2 \sin^{2}(\theta)}} =\frac{\pi}{2} \left(1+ \left(\frac{1}{2}\right)^2+ \left(\frac{1.3}{2.4}\right)^2+ \cdots \right) \end{eqnarray*} & it is well known that this can be evaluated for special values of $k$ in terms of Gamma functions whose arguements are rational values. See this question Can $\Gamma(1/5)$ be written in this form? and the reference cited there. So my first thought is to use $(1)$ three times, sum the geometric plum & we have \begin{eqnarray*} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \frac{d \alpha d \beta d \gamma}{1+ \sin^2 \alpha \sin^2 \beta \sin^2 \gamma} \end{eqnarray*} This triple integral looks difficult so ...

Second thoughts ... use $(1)$ twice & then use $(2)$ to get the double integral \begin{eqnarray*} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \frac{d \alpha d \beta }{\sqrt{1+ \sin^2 \alpha \sin^2 \beta }} \end{eqnarray*} now substitute $ \sqrt{s} =\sin \alpha$ and $ \sqrt{t} =\sin \beta$ (might have lost a factor of $4$) \begin{eqnarray*} \int_{0}^{1} \int_{0}^{1} \frac{d s d t }{\sqrt{s(1-s)t(1-t)(1+st) }} \end{eqnarray*} & I am not sure what to do with this.

In both of these attempts I feel I have taken a wrong turn.

Can someone either give me a Big hint or a reference to the original derivation of this result or a reasonably complete solution ?

Bonus question ... why was it stated with $\frac{10}{8}$ instead of $\frac{5}{4}$ ?

The result $$1-\left( \frac{1}{2} \right)^{3}+\left( \frac{1\cdot 3}{2\cdot 4} \right)^{3}-\left( \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6} \right)^{3}+\left( \frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8} \right)^{3}-...=\frac{\Gamma ^{2}\left( \frac{9}{8} \right)}{\Gamma ^{2}\left( \frac{7}{8} \right)\cdot \Gamma ^{2}\left( \color{red}{\frac{5}{4}} \right)}$$

Hint from Jack D'Auirzio : look at equation (6) here http://mathworld.wolfram.com/CompleteEllipticIntegraloftheFirstKind.html and use $2kk'=i$ where $k'$ is the complementary modulus.

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    $\begingroup$ Extended hint: once $2kk'=i$ is solved, it is enough to convert the LHS of $(6)$ into a value of Euler's Beta function. And if someone wonders about how to prove $(6)$, that can be shown through different techniques: differential equations, Landen's transforms for hypergeometric functions, relations between elliptic integrals of the first kind and the AGM. $\endgroup$ – Jack D'Aurizio Aug 14 '17 at 23:20
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    $\begingroup$ They probably used $10/8$ to keep the pattern of $n/8$ going in those Gamma functions. $\endgroup$ – Simply Beautiful Art Aug 14 '17 at 23:38
  • $\begingroup$ Related: math.stackexchange.com/a/93878/44121 $\endgroup$ – Jack D'Aurizio Aug 15 '17 at 0:41
  • $\begingroup$ +1 for adding all that context to a poorly framed question. $\endgroup$ – Paramanand Singh Aug 15 '17 at 2:06
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    $\begingroup$ $$\sum_{k=0}^{\infty} \frac{(2n-1)!!}{(2n)!!} y^n = \frac{1}{\sqrt{1-y}}$$, Shouldn't it be $n = 0$ as the index ? $\endgroup$ – A---B Sep 23 '17 at 12:23

The sum evaluates to $(1-k^{2})(2K(k)/\pi)^{2}$ for $k=\sqrt{2}-1$. The value of $K$ is obtained from here. Using the value of $K$ we can get the desired closed form sum of the given series.

Using hypergeometric transformations we can prove the formula $$\left(\frac{2K(k)}{\pi}\right)^{2}=\sum_{n=0}^{\infty}\left(\frac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdot 6\cdots (2n)}\right)^{3}(2kk')^{2n}\tag{1}$$ (see complete proof here). Note that the above equation can be thought of as an identity involving the nome $q=e^{-\pi K'/K} $ also and then we can switch from $q$ to $-q$. In terms of Jacobi's theta functions we have $2K/\pi=\vartheta_{3}^{2}(q)$ and replacing $q$ by $-q$ leads us to $$\vartheta_{3}^{2}(-q)=\vartheta_{4}^{2}(q)=\frac{\vartheta_{4}^{2}(q)}{\vartheta_{3}^{2}(q)}\cdot\vartheta_{3}^{2}(q)=k'\cdot\frac{2K}{\pi}$$ Similarly we have $$(2kk')^{2}=4k^{2}k'^{2}=4\cdot\frac{\vartheta_{2}^{4}(q)\vartheta_{4}^{4}(q)}{\vartheta_{3}^{8}(q)}=4\cdot\frac{16q\psi^{4}(q^{2})\vartheta_{4}^{4}(q)}{\vartheta_{3}^{8}(q)}$$ where $\psi$ is one of Ramanujan's theta functions. Now changing $q$ into $-q$ interchanges $\vartheta_{3},\vartheta_{4}$ and hence the above expression becomes $$-4\cdot\frac{16q\psi^{4}(q^{2})\vartheta_{3}^{4}(q)}{\vartheta_{4}^{8}(q)} =-4\cdot\frac{\vartheta_{2}^{4}(q)/\vartheta_{3}^{4}(q)} {\vartheta_{4}^{8}(q)/\vartheta_{3}^{8}(q)}=-\frac{4k^{2}}{k'^{4}} $$ So we have the formula $$k'^{2}\left(\frac{2K}{\pi}\right)^{2}=\sum_{n=0}^{\infty}(-1)^{n}\left(\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots(2n)}\right)^{3}(2k/k'^{2})^{2n}\tag{2}$$ And putting $k=\sqrt{2}-1$ we get the desired sum.

As indicated in a comment from Jack d'Aurizio the sum in question can also be expressed in terms of arithmetic-geometric-mean as $$\frac{2}{\operatorname {AGM} (\sqrt{2},\sqrt{1+\sqrt{2}})^{2}}\tag{3}$$ Writing in this manner helps to evaluate the sum very accurately with a few iterations of the AGM.

It is also interesting to know that Ramanujan used the series $(1)$ to obtain $$\frac{4}{\pi}=\sum_{n=0}^{\infty}\left(\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}\right)^{3}\cdot\frac{6n+1}{4^{n}}\tag{4}$$ and $$\frac{2}{\pi}=\sum_{n=0}^{\infty}(-1)^{n}(4n+1)\left(\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}\right)^{3}\tag{5}$$ was obtained from series $(2)$. You can get all the details here.

The value of $K(\sqrt{2}-1)$ has been obtained via non-obvious hypergeometric transformations and it would be nice to have an approach based on direct evaluation of integral involved.

  • $\begingroup$ (+1) Beautiful answer, I was waiting for Clausen's formula to appear. It might be an interesting addendum to mention that the origin series equals $$ \frac{2}{\text{AGM}(\sqrt{2},\sqrt{1+\sqrt{2}})^2}.$$ $\endgroup$ – Jack D'Aurizio Aug 15 '17 at 2:12
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    $\begingroup$ @JackD'Aurizio: I will add your comment in my answer (because at times people might feel a bit lazy to read comments). And thanks for your kind words. $\endgroup$ – Paramanand Singh Aug 15 '17 at 3:08
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    $\begingroup$ \begin{eqnarray*} \Gamma \left( \frac{1}{8} \right) =\pi ^{\frac{1}{8}} 2^{\frac{17}{8}} K \left( \frac{1}{\sqrt{2}} \right)^{\frac{1}{4}} K \left( \sqrt{2}-1 \right)^{\frac{1}{2}} \end{eqnarray*} I will read the two references in your post. I think I can piece it together from here. Thank you very much indeed for your efforts. $\endgroup$ – Donald Splutterwit Aug 15 '17 at 20:57
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    $\begingroup$ @JackD'Aurizio Thank you very much for helpful hints. $\endgroup$ – Donald Splutterwit Aug 15 '17 at 20:58

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