How can I prove that $\sum_{n=0}^\infty\frac{(2n)!}{2^{2n}(n!)^2(2n+1)}$ converges? Note: If this is a duplicate question, and I'm pretty sure it is, I have not been able to find a post that it duplicates. The closest one I could find asks about the divergent series
$$\sum_{n=0}^\infty \frac{(2n)!}{2^{2n}(n!)^2}$$
Now for the problem: I'd like to prove that the series
$$\sum_{n=0}^\infty\frac{(2n)!}{2^{2n}(n!)^2(2n+1)}$$
converges.
This series is the formal result when one evaluates the Maclaurin series of $\sin^{-1}$ at $1$. Appealing to Abel's theorem and consulting WolframAlpha, the series, if convergent, sums to $\pi/2$. This is to be expected because $\sin^{-1}(1)=\pi/2$.
Given the presence of factorials and exponentials, my first idea was to try and apply the ratio test to the terms of the series. Unfortunately, the relevant limit evaluates to $1$:
\begin{align}
\frac{\frac{(2(n+1))!}{2^{2(n+1)}((n+1)!)^2(2(n+1)+1)}}{\frac{(2n)!}{2^{2n}(n!)^2(2n+1)}} &= \frac{(2n+2)!}{2^{2n+2}((n+1)n!)^2(2n+3)}\cdot\frac{2^{2n}(n!)^2(2n+1)}{(2n)!}\\
&= \frac{(2n)!(2n+1)(2n+2)}{2^2(n+1)^2(n!)^2(2n+3)}\cdot\frac{(n!)^2(2n+1)}{(2n)!}\\
&= \frac{(2n+1)^2\cdot 2(n+1)}{2^2(n+1)^2(2n+3)}\\
&= \frac{4n^2+4n+1}{2(n+1)(2n+3)}\\
&= \frac{4n^2+4n+1}{4n^2+10n+6}\to 1\text{ as }n\to\infty\\
\end{align}
It is pointless to apply the root test since it is inconclusive whenever the ratio test is. I would consider trying to leverage the identity
$$\frac{(2n)!}{2^{2n}(n!)^2}=\frac{2}{\pi}\int_0^\frac{\pi}{2}\sin^{2n}(x)dx$$
but this requires knowledge of the decay rate of $\int_0^\frac{\pi}{2}\sin^{2n}(x)dx$, something I do not yet possess. I doubt integration by parts will be useful, since this is how you prove the integral identity, and obvious substitutions like $x=\sin^{-1}(t)$ yield seemingly non-fruitful expressions like
$$\frac{2}{\pi}\int_0^1\frac{x^{2n}}{\sqrt{1-x^2}}dx$$
I don't have any other ideas. Could I get some assistance?
 A: To elaborate on KStarGamer's answer. Raabe's test tells you that if
$$lim_{n\rightarrow \infty} n(1-\frac{a_{n+1}}{a_n}) = L, \hspace{2mm} a_n\geq0$$
then, $$\text{If} \hspace{2mm}L>1, \hspace{2mm}\sum a_{n} \hspace{2mm}\text{converges.}$$
$$\text{If} \hspace{2mm}L<1, \hspace{2mm}\sum a_{n} \hspace{2mm}\text{diverges.}$$
If you want to look it up, Raabe Test.
On your example, after getting $\frac{a_{n+1}}{a_n} =\frac{4n^2+4n+1}{4n^2+10n+6}$ you substitute it on the limit above and you get,
$$lim_{n\rightarrow \infty}n(1-\frac{4n^2+4n+1}{4n^2+10n+6})=lim_{n\rightarrow \infty } \frac{6n^2+5n}{4n^2+10n+6} = \frac{3}{2}>1$$
So, thanks to Raabe's test we have shown that $\sum_{n}a_{n}<\infty$, i.e. converges.
A: Here is another possibility.  First, verify the following inequality:
$$
\frac{1}{2\sqrt{n}} \le \frac{(2n)!}{2^{2n} (n!)^2} \le \frac{1}{\sqrt{2n}}.
$$
You can prove this by induction.  Then for your series you have
$$
\frac{1}{2\sqrt{n}(2n+1)} \le \frac{(2n)!}{2^{2n} (n!)^2 (2n+1)} \le \frac{1}{\sqrt{2n} (2n+1)},
$$
so your series converges by the comparison test.
