TITLE: How to prove by arithmetical means that $\sum\limits_{k=1}^\infty \frac{((k-1)!)^2}{(2k)!} =\frac{1}{3}\sum\limits_{k=1}^{\infty}\frac{1}{k^{2}}$
QUESTION [17 upvotes]: I've been trying to prove, by arithmetical means, that
$$\sum_{k=1}^\infty \frac{((k-1)!)^2}{(2k)!} =\frac{1}{3}\sum_{k=1}^{\infty}\frac{1}{k^{2}}$$
without success.
When I say "by arithmetical means" I mean to say, go from the left to the right expression just by symbolic manipulation.
Can anyone devise a way of doing this?

REPLY [14 votes]: First, let's compute a slightly simpler sum
$$
\begin{align}
\sum_{k=1}^\infty\frac{\Gamma(k)^2}{\Gamma(2k)}(2x)^{2k-1}
&=\sum_{k=1}^\infty\mathrm{B}(k,k)(2x)^{2k-1}\\
&=\sum_{k=1}^\infty\int_0^1t^{k-1}(1-t)^{k-1}(2x)^{2k-1}\mathrm{d}t\\
&=\int_0^1\frac{2x}{1-4x^2t(1-t)}\mathrm{d}t\\
&=\frac{1}{2x}\int_0^1\frac{1}{t^2-t+\frac{1}{4x^2}}\mathrm{d}t\\
&=\frac{1}{2x}\int_0^1\frac{1}{(t-\alpha)(t-1+\alpha)}\mathrm{d}t\text{ where }2\alpha-1=\sqrt{1-\frac{1}{x^2}}\\
&=\frac{1}{2x}\frac{1}{2\alpha-1}\int_0^1\left(\frac{1}{t-\alpha}-\frac{1}{t-1+\alpha}\right)\mathrm{d}t\\
&=\frac{1}{2\sqrt{x^2-1}}\left[\log\left(\frac{\alpha-1}{\alpha}\right)-\log\left(\frac{\alpha}{\alpha-1}\right)\right]\\
&=\frac{1}{\sqrt{x^2-1}}\log\left(\frac{\alpha-1}{\alpha}\right)\text{ where }\frac{\alpha-1}{\alpha}=\left(\sqrt{1-x^2}+ix\right)^2\\
&=\frac{-2i}{\sqrt{1-x^2}}i\,\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\\
&=\frac{2}{\sqrt{1-x^2}}\sin^{-1}(x)\tag{1}
\end{align}
$$
Integrating both sides of $(1)$ yields
$$
\frac12\sum_{k=1}^\infty\frac{\Gamma(k)^2(2x)^{2k}}{\Gamma(2k+1)}
=\left[\sin^{-1}(x)\right]^2\tag{2}
$$
Plugging $x=\frac12$ into $(2)$ gives
$$
\sum_{k=1}^\infty\frac{(k-1)!^2}{(2k)!}=\frac{\pi^2}{18}\tag{3}
$$
In this answer, it is shown that
$$
\sum_{k=1}^\infty\frac{1}{k^2}=\zeta(2)=\frac{\pi^2}{6}\tag{4}
$$
Combining $(3)$ and $(4)$ yields
$$
\sum_{k=1}^\infty\frac{(k-1)!^2}{(2k)!}=\frac13\sum_{k=1}^\infty\frac{1}{k^2}\tag{5}
$$