Series of square of factorial I recently found this series and the result:
$$
  \sum_{k=1}^{\infty}{\frac{((k-1)!)^2}{(2k)!}} = \frac{\pi^2}{18}
$$
However I'm not able to justify why. Is there a special function (zeta, beta) that I don't know and that may help?
Thank you
for an even $k$, $(2k)!=2^kk!(2k-1)!!$ which may help, also notice that:
$$\sum_{k=1}^\infty\frac{((k-1)!)^2}{(2k)!}=\sum_{k=1}^\infty\frac{(k!)^2}{k^2(2k)!}$$
also since:
$$k!=\prod_{n=1}^kn,(2k)!=\prod_{n=1}^{2k}n\Rightarrow\frac{k!}{(2k)!}=\prod_{n=k+1}^{2k}\frac 1n=\prod_{n=1}^k\frac{1}{n+k}$$
and so:
$$\frac{(k!)^2}{(2k)!}=\prod_{n=1}^k\frac{n}{n+k}$$
so we can rewrite our expression as:
$$S=\sum_{k=1}^\infty\prod_{n=1}^k\frac{n}{k^2(n+k)}$$
now we should try and do something with this product:
$$\frac{n}{n+k}=1-\frac{k}{n+k}$$

$$\ln\left(\frac{n}{n+k}\right)=\sum_{j=1}^\infty\frac{(-1)^j n^j}{j(n+k)^j}$$
although I'm not sure where to go from here
for an even $k$, $(2k)!=2^kk!(2k-1)!!$ which may help, also notice that:
$$\sum_{k=1}^\infty\frac{((k-1)!)^2}{(2k)!}=\sum_{k=1}^\infty\frac{(k!)^2}{k^2(2k)!}$$
also since:
$$k!=\prod_{n=1}^kn,(2k)!=\prod_{n=1}^{2k}n\Rightarrow\frac{k!}{(2k)!}=\prod_{n=k+1}^{2k}\frac 1n=\prod_{n=1}^k\frac{1}{n+k}$$
and so:
$$\frac{(k!)^2}{(2k)!}=\prod_{n=1}^k\frac{n}{n+k}$$
so we can rewrite our expression as:
$$S=\sum_{k=1}^\infty\prod_{n=1}^k\frac{n}{k^2(n+k)}$$
now we should try and do something with this product:
$$\frac{n}{n+k}=1-\frac{k}{n+k}$$

$$\ln\left(\frac{n}{n+k}\right)=\sum_{j=1}^\infty\frac{(-1)^j n^j}{j(n+k)^j}$$
although I'm not sure where to go from here