TITLE: Compute: $\sum\limits_{n=1}^{\infty}\frac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)\cdot (2n+1)}$
QUESTION [18 upvotes]: 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!

REPLY [8 votes]: We have 
$$\sum_{n=1}^{\infty}\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)\cdot (2n+1)} 
= \sum_{n=1}^\infty {2n\choose n}\frac{1}{4^n(2n+1)}.$$
Now notice that 
$$\int_0^{1/2}dx\, 2x^{2n} = \frac{1}{4^n(2n+1)}.$$
Thus, we need only compute the sum 
$$\begin{eqnarray*}
2\sum_{n=1}^\infty {2n\choose n}x^{2n}
&=& 2\sum_{n=0}^\infty {2n\choose n}x^{2n} - 2 \\
&=& \frac{2}{\sqrt{1-4x^2}} - 2
\end{eqnarray*}$$
and integrate.

REPLY [8 votes]: Let's rewrite this as :
$$\sum_{n=1}^{\infty}\frac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)\cdot (2n+1)}=\sum_{n=1}^{\infty}\frac {(2n)!} {4^n (n!)^2(2n+1)}$$
But this is merely the expansion of $\arcsin(x)-1$ for $x=1$ as found at this page on central binomial series :
$$\arcsin(x)=2\sum_{n=0}^{\infty}\frac{\binom{2n}{n}}{2n+1}\left(\frac x2\right)^{2n+1}$$
That is $\ \displaystyle \arcsin(1)-1=\frac {\pi}2-1$
The previous link to Boris Gourévitch work concerning $\pi$ is rather interesting because of the many relations provided concerning $\binom{2n}{n}$ and because it shows that the central binomial term may be at the numerator as well as at the denominator (see $(378)$).
It will be at the numerator for $\arcsin(x)$ and at the denominator for $\arcsin(x)^2$ and $\dfrac{\arcsin(x)}{\sqrt{1-x^2}}$ (relation $(396)$) !