Compute: $\sum\limits_{n=1}^{\infty}\frac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)\cdot (2n+1)}$ Compute the sum:
$$\sum_{n=1}^{\infty}\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)\cdot (2n+1)}$$
At the moment, I only know that it's convergent and this is not hard to see if you look at the answers here I received for other problem with a similar series. For the further steps I need
some hints if possible. Thanks!

Starting with the power series derived using the binomial theorem,
$$
(1-x)^{-1/2}=1+\tfrac12x+\tfrac12\tfrac32x^2/2!+\tfrac12\tfrac32\tfrac52x^3/3!+\dots+\tfrac{1\cdot2\cdot3\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}x^n+\cdots
$$
and integrating, we get the series for
$$
\sin^{-1}(x)=\int_0^x(1-t^2)^{-1/2}\mathrm{d}t=\sum_{n=0}^\infty\tfrac{1\cdot2\cdot3\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\frac{x^{2n+1}}{2n+1}
$$
Setting $x=1$, we get
$$
\sum_{n=1}^\infty\tfrac{1\cdot2\cdot3\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\frac{1}{2n+1}=\sin^{-1}(1)-1=\frac\pi2-1
$$
