Find $\sum_{n=1}^{\infty} \frac {1^2\times3^2\times...\times(2n-1)^2}{2^{2n}(2n+1)!}=?$
So my guess is this is some kind of $\arcsin(x)$.
My attempt:
Let $f(x) = \arcsin(x)$ then
$f'(x) = \frac 1{\sqrt{1-x^2}}$
We know that : $(1+x)^{\alpha}=1+\sum_{n=1}^{\infty}\frac{\alpha(\alpha-1)...(\alpha-n+1)}{n!}x^n$.So then with $\alpha=-\frac 12$ and $x\leftarrow-x^2$ we get that:
$$\frac 1{\sqrt{1-x^2}}=1+\sum_{n=1}^{\infty} \frac {(-\frac12)(-\frac 32)(-\frac 52)...(-\frac {2n-1}2)}{n!}(-1)^nx^{2n}=1+\sum_{n = 1}^{\infty}\frac{1\times3\times5\times...\times(2n-1)}{2^nn!}x^{2n}.$$
if we integrate both sides we get that:
$$f(x)=x+\sum_{n=1}^{\infty}\frac {1\times3\times...\times(2n-1)}{2^nn!(2n+1)}x^{2n+1}$$
Is there something wrong with my expansion? I think I might've made a mistake somewhere
$$\sum_{n\geq 1}\frac{(2n-1)!!^2}{4^n(2n+1)!}=\sum_{n\geq 1}\frac{\binom{2n}{n}}{16^n (2n+1)}=\int_{0}^{1}\left(\frac{2}{\sqrt{4-x^2}}-1\right)\,dx=2\arcsin\frac{1}{2}-1=\frac{\pi}{3}-1. $$