<s>[INST] Why do the endpoints of the Maclaurin series for arcsin converge? 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. 
 [/INST] Here's another approach. 
We write the shorthand $m!$ for $\Gamma(m+1)$, where $\Gamma$ is the gamma function.
Let $a_n = \displaystyle\frac{(-1)^n}{2n+1} {-\frac{1}{2} \choose n}$. 
Then 
$$\begin{eqnarray*}
\frac{a_{n+1}}{a_n} &=& -\frac{2n+1}{2n+3} \,
\frac{{-\frac{1}{2}\choose n+1}}{{-\frac{1}{2}\choose n}} \\
&=& -\frac{2n+1}{2n+3} \,
\frac{(-\frac{1}{2})!}{(n+1)!(-\frac{1}{2}-n-1)!}\,
\frac{n!(-\frac{1}{2}-n)!}{(-\frac{1}{2})!} \\
&=& -\frac{2n+1}{2n+3}\,
\frac{-\frac{1}{2}-n}{n+1} \\
&=& \frac{(2n+1)^2}{2(n+1)(2n+3)} \\
&=& 1-\frac{3}{2n} +O\left(\frac{1}{n^2}\right).  
\end{eqnarray*}$$
Since $3/2>1$ the series converges by Raabe's test. 
(Note that the $a_n$s are positive, so we don't bother taking the absolute value.)
</s>