A series inspired by Satyajit

If u have a different solution to it, plz do post.

As mentioned that this problem is inspired by @Satyajit Mohanty

$\large{(\arcsin x)^2=\displaystyle \sum^{\infty}_{n=0} \frac{2^{2n} (n!)^{2}}{(2n+1)! (n+1)} x^{2n+2}}$ ................... $(*)$

The given series $nth$ term is

Note that $.$ means $\times$

$\large{T_{n}=\frac{2.4.6........2n}{3.5.7....(2n+1)(2n+2)}}$

$\large{\displaystyle \sum^{\infty}_{n=1}T_{n}=\displaystyle \sum^{\infty}_{n=1} \frac{2^{2n} (n!)^{2}}{(2n+2)!}}$

On rearranging $(*)$ and replacing $x$ by $1)$

$\large{\frac{1}{2}(\arcsin 1)^2=\displaystyle \sum^{\infty}_{n=0} \frac{2^{2n} (n!)^{2}}{(2n+2)!}}$

$\large{\frac{1}{2}(\arcsin 1)^2=\frac{1}{2}+\displaystyle \sum^{\infty}_{n=1} \frac{2^{2n} (n!)^{2}}{(2n+2)!}}$

Thus

$\large{\displaystyle \sum^{\infty}_{n=1} \frac{2^{2n} (n!)^{2}}{(2n+2)!}=\frac{\pi^{2}}{8}-\frac{1}{2}}$

I have a different solution, actually not much different.

It is for those who don't know that ${{sin}^{-1}(x)}^{2}$ series(including me).

So,

$\displaystyle T = \sum_{n=1}^{\infty}{\frac{(2n)!!}{(2n+1)!!(2n+2)}}$ is our problem

Let

$\displaystyle S(x) = \sum_{n=1}^{\infty}{\frac{(2n)!!}{(2n+1)!!} x^n}$

$\displaystyle S(x) = \sum_{n=1}^{\infty}{\frac{{2}^{2n} n!^2}{(2n+1)!} x^n}$

$\displaystyle S(x) = \sum_{n=1}^{\infty}{\frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+1)} {4x}^{n}}$

Beta function, aha?

$\displaystyle S(x) = \sum_{n=1}^{\infty}{\text{B}(n,n) {4x}^{n}}$

$\displaystyle S(x) = \sum_{n=1}^{\infty}{\int_{0}^{1}{t^n{(1-t)}^{n} \ dt} {4x}^{n}}$

$\displaystyle S(x) = \int_{0}^{1}{\frac{4xt(1-t)}{1-4xt(1-t)} \ dt}$

$\displaystyle S(x) = \int_{0}^{1}{\frac{dt}{4xt^2 - 4xt + 1}} - 1$

$\displaystyle S(x) = \frac{{tan}^{-1}\left(\frac{x}{\sqrt{x-x^2}}\right)}{\sqrt{x-x^2}} - 1$ [I'd not be giving details how to solve that integral. Probably that's what they say "Left as an exercise for the reader!"]

$\displaystyle 2T = \int_{0}^{1}{S(x) \ dx}$ , Right?

$\displaystyle T = \frac{1}{2} \int_{0}^{1}{\frac{{\tan}^{-1}\left(\frac{x}{\sqrt{x-x^2}}\right)}{\sqrt{x-x^2}} - 1 \ dx}$

$\displaystyle T = \frac{1}{2} \int_{0}^{1}{\frac{{\tan}^{-1}\left(\frac{x}{\sqrt{x-x^2}}\right)}{\sqrt{x-x^2}} \ dx} - \frac{1}{2}$

Well, as I am lazy enough to not provide a solution to this integral as well, it is again left as an exercise for the reader :P(so many exercises! Well, I may give solutions later if required but these are elementary, I think exercise would be fine.)

Computation shows that it evaluates to

$\displaystyle T = \frac{\pi^2}{8} - \frac{1}{2}$