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. 
 A: Method 1:
First, we have
$$
\frac{(-1)^n}{2n+1}\binom{-1/2}{n}=\frac{1}{2n+1}\binom{n-1/2}{n}\tag{1}
$$
and
$$
\binom{n-1/2}{n}=\frac{\Gamma(n+1/2)}{\Gamma(n+1)\Gamma(1/2)}=\frac{\Gamma(n+1/2)}{\Gamma(n+1)\sqrt{\pi}}\tag{2}
$$
By Gautschi's Inequality, we have that
$$
\frac{1}{\sqrt{n+1}}\le\frac{\Gamma(n+1/2)}{\Gamma(n+1)}\le\frac{1}{\sqrt{n}}\tag{3}
$$
Therefore, we get that
$$
\frac{1}{(2n+1)\sqrt{\pi}}\frac{1}{\sqrt{n+1}}\le\frac{(-1)^n}{2n+1}\binom{-1/2}{n}\le\frac{1}{(2n+1)\sqrt{\pi}}\frac{1}{\sqrt{n}}\tag{4}
$$
By comparison to
$$
\sum_{n=1}^\infty\frac{1}{n^{3/2}}
$$
We get that the series converges.

Method 2:
As Gerry Myerson suggests, Stirling's Approximation gives
$$
\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}\tag{5}
$$
Thus the term in your third form is
$$
\frac{(2n)!}{2^{2n}(n!)^2}\frac{1}{2n+1}\sim\frac{1}{\sqrt{\pi n}}\frac{1}{2n+1}\tag{6}
$$
Again, comparison to
$$
\sum_{n=1}^\infty\frac{1}{n^{3/2}}
$$
yields the convergence of the series.
A: For $n \geq 1$ let
$$f_n(x) = \frac{1}{2n - 1} + \frac{2n - 1}{2n} x
$$
Then the $N$-th partial sum $S_N$ is
$$
S_N = \sum_{n=0}^N\frac{1}{2} \frac{3}{4} \cdots \frac{2n - 1}{2n} \frac{1}{2n + 1} = f_1 \circ f_2 \circ \cdots \circ f_N \left( \frac{1}{2N+1} \right)
$$
Each $f_n$ maps the interval $[0, 2]$ into itself and therefore all partial sums are bounded above by $2$.
A: 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.)
A: 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.
