TITLE: Why do the endpoints of the Maclaurin series for arcsin converge?
QUESTION [7 upvotes]: The series $$\sum_{n=0}^\infty {{-\frac {1} 2} \choose n} \frac{(-1)^n}{2n+1}$$ is an endpoint for the Maclaurin series for arcsin(x). (The other endpoint is just the negative of this one.) I played around with this a bit and turned it into three (potentially useful) forms :
$\sum_{n=0}^\infty \frac12\frac34\frac56\cdots\frac{2n-1}{2n} \frac{1}{2n+1}$
$\sum_{n=0}^\infty \left( 1-\frac 1 2 \right) \left( 1-\frac 1 4 \right)\cdots \left( 1-\frac 1 {2n} \right)\frac{1}{2n-1}$
$\sum_{n=0}^\infty \dfrac{(2n)!}{2^{2n}(n!)^2}\dfrac{1}{2n+1}$, 
but I'm not sure where one could go from here.

REPLY [2 votes]: Another approach would be Stirling's closedly related Wallis' formula. It gives
$$\lim_{n \to \infty} \frac{(2n)!}{(2n-1)!}\frac{1}{\sqrt n}=\sqrt{\pi}$$
This means that your $n$th term is asymptotically equal to  
$$ {\sqrt{\frac 1 {\pi n}}}\frac{1}{2n+1}$$
which explains why the series converges. In particular, 
$$\lim_{n \to \infty} \frac{n^{3/2}}{\sqrt{n}(2n+1)}=\frac 1 2 $$ so the convergence of $$\sum_{n >0}  n^{-3/2} $$  implies that of your series.