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?

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
$$\sum_{n=1}^\infty a_n\,x^n=\frac{\sin ^{-1}(x)^2-x^2}{2 x^2}$$
$$\sum_{n=0}^\infty b_n\,x^n=\frac{\sin ^{-1}(x)-x}{x^3}$$
