Summation of $n$th partial products of the square of even numbers diverges, but for odd numbers they converge in this series I'm looking at. Why? So I have the two following series:
$$\sum_{n=1}^\infty \frac{\prod_{k=1}^n(2k)^2}{(2n+2)!}$$
$$\sum_{n=0}^\infty \frac{\prod_{k=0}^n(2k+1)^2}{(2n+3)!}$$
I figured out the $n$th partial products:
$$\prod_{k=1}^n(2k)^2=4^n(n!)^2$$
$$\prod_{k=0}^n (2k+1)^2=\frac{((2n+1)!)^2}{4^n(n!)^2}$$
So putting these back into my series they become the following:
$$\sum_{n=1}^\infty \frac{\prod_{k=1}^n(2k)^2}{(2n+2)!}=\sum_{n=1}^\infty\frac{4^n(n!)^2}{(2n+2)!}$$
Now this diverges as expected by the limit test test. However when I look at my other series:
$$\sum_{n=0}^\infty \frac{\prod_{k=0}^n(2k+1)^2}{(2n+3)!}=\sum_{n=0}^\infty\frac{(
(2n+1)!)^2}{4^n(n!)^2(2n+3)!}$$
By the limit test maybe diverges or maybe doesn't, and the ratio test is inconclusive. Since I wasn't sure what to use for the a comparison test I threw this into wolfram alpha and it told me it converges which is baffling to me since both series are very similar if we write them out:
$$\sum_{n=1}^\infty \frac{\prod_{k=1}^n(2k)^2}{(2n+2)!}=\frac{2^2}{4!}+\frac{2^24^2}{6!}+\frac{2^24^26^2}{8!}\cdot\cdot\cdot\cdot$$
$$\sum_{n=0}^\infty \frac{\prod_{k=0}^n(2k+1)^2}{(2n+3)!}=\frac{1^2}{3!}+\frac{1^23^2}{5!}+\frac{1^23^25^2}{7!}+\cdot\cdot\cdot$$
They both have the nth parial product of the even/odd integers squared in the numerator, and are over a factorial that is two greater than $n$, so I'm not sure why one is diverging and the other is converging. Is wolframalpha wrong, as it can be at times? Or is there someething here that I am missing?
 A: Elaborating after @Erick Wong's comments.
You properly found that
$$a_n=\frac{4^n(n!)^2}{(2n+2)!}$$ Take logarithms
$$\log(a_n)=n \log(4)+2\log(n!)-\log((2n+2)!)$$ Use Stirling approximation twice and continue with Taylor series to find
$$\log(a_n)=\left(\frac{3}{2} \log \left(\frac{1}{n}\right)+\log \left(\frac{\sqrt{\pi
   }}{4}\right)\right)-\frac{11}{8 n}+O\left(\frac{1}{n^2}\right)$$ that is to say
$$a_n \sim \frac{\sqrt \pi}{4 n^{\frac 32}}\exp\left(-\frac{11}{8 n}\right) <\frac{\sqrt \pi}{4 n^{\frac 32}}$$
$$\sum_{n=1}^\infty \frac{\sqrt \pi}{4 n^{\frac 32}}=\frac{\sqrt{\pi }}{4}  \zeta \left(\frac{3}{2}\right)\approx 1.15758$$
Sooner or later, you will learn that
$$\sum_{n=1}^\infty \frac{4^n(n!)^2}{(2n+2)!}=\frac{\pi ^2-4}{8}\approx 0.73370$$
Doing the same with
$$b_n=\frac{(2n+1)!^2}{4^n(n!)^2(2n+3)!}$$
$$\log(b_n)=2\log((2n+1)!)-n \log(4)-2\log(n!)-\log((2n+3)!)$$
$$\log(b_n)=\left(\frac{3}{2} \log \left(\frac{1}{n}\right)+\log \left(\frac{1}{2 \sqrt{\pi
   }}\right)\right)-\frac{17}{8 n}+O\left(\frac{1}{n^2}\right)$$  that is to say
$$b_n \sim \frac{1}{2 \sqrt \pi n^{\frac 32}}\exp\left(-\frac{17}{8 n}\right) < \frac{1}{2 \sqrt \pi n^{\frac 32}}$$
$$\sum_{n=1}^\infty \frac{1}{2 \sqrt \pi n^{\frac 32}}=\frac{1}{2 \sqrt \pi }\zeta \left(\frac{3}{2}\right)\approx 0.73694$$
Sooner or later, you will learn that
$$\sum_{n=0}^\infty \frac{(2n+1)!^2}{4^n(n!)^2(2n+3)!}=\frac{ \pi -2}{2} \approx 0.57080$$
Edit
Notice that
$$\sum_{n=1}^\infty a_n\,x^n=\frac{\sin ^{-1}(x)^2-x^2}{2 x^2}$$
$$\sum_{n=0}^\infty b_n\,x^n=\frac{\sin ^{-1}(x)-x}{x^3}$$
A: Convergence
Using the asymptotic approximation given in inequality $(9)$ of this answer, we get
$$
\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}\tag1
$$
Therefore,
$$
\begin{align}
\frac{\prod\limits_{k=1}^n(2k)^2}{(2n+2)!}
&=\frac{4^nn!^2}{(2n)!(2n+1)(2n+2)}\\
&=\frac{\color{#090}{4^n}}{\color{#090}{\binom{2n}{n}}\color{#C00}{(2n+1)(2n+2)}}\\
&\sim\frac{\color{#090}{\sqrt{\pi n}}}{\color{#C00}{4n^2}}\\
&=\frac{\sqrt\pi}{4}\frac1{n^{3/2}}\tag2
\end{align}
$$
and
$$
\begin{align}
\frac{\prod\limits_{k=0}^n(2k+1)^2}{(2n+3)!}
&=\frac{(2n+1)!^2}{4^nn!^2(2n+3)!}\\
&=\frac{\color{#090}{\binom{2n}{n}}\color{#C00}{(2n+1)}}{\color{#090}{4^n}\color{#C00}{(2n+2)(2n+3)}}\\
&\sim\frac1{\color{#090}{\sqrt{\pi n}}\,\color{#C00}{2n}}\\
&=\frac1{2\sqrt\pi}\frac1{n^{3/2}}\tag3
\end{align}
$$
The sums of both $(2)$ and $(3)$ converge by comparison to a $p$-series with $p=3/2$.

Evaluation
In this answer, it is shown that
$$
\begin{align}
\arcsin^2(x)
&=\sum_{k=1}^\infty\frac{4^kx^{2k}}{2k^2\binom{2k}{k}}\\
&=\sum_{k=1}^\infty\frac{4^k}{\binom{2k}{k}}\frac{x^{2k}}{2k^2}\\
&=\sum_{k=0}^\infty\frac{4^k}{\binom{2k}{k}}\frac{2x^{2k+2}}{(2k+1)(2k+2)}\tag4\\
\end{align}
$$
and in this answer, it is shown that
$$
\begin{align}
\arcsin(x)
&=\sum_{k=0}^\infty\frac2{2k+1}\binom{2k}{k}\left(\frac{x}{2}\right)^{2k+1}\\
&=\sum_{k=0}^\infty\frac{\binom{2k}{k}}{4^k}\frac{x^{2k+1}}{2k+1}\\
&=x+\sum_{k=0}^\infty\frac{\binom{2k}{k}}{4^k}\frac{(2k+1)x^{2k+3}}{(2k+2)(2k+3)}\tag5
\end{align}
$$
Applying $(4)$, we get
$$
\begin{align}
\sum_{n=1}^\infty\frac{\prod\limits_{k=1}^n(2k)^2}{(2n+2)!}
&=\sum_{n=1}^\infty\frac{4^n}{\binom{2n}{n}(2n+1)(2n+2)}\\
&=\frac12\arcsin(1)^2-\frac12\\
&=\frac{\pi^2}8-\frac12\tag6
\end{align}
$$
Applying $(5)$, we get
$$
\begin{align}
\sum_{n=0}^\infty\frac{\prod\limits_{k=0}^n(2k+1)^2}{(2n+3)!}
&=\sum_{n=0}^\infty\frac{\binom{2n}{n}(2n+1)}{4^n(2n+2)(2n+3)}\\
&=\arcsin(1)-1\\[6pt]
&=\frac\pi2-1\tag7
\end{align}
$$
