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?

Question

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?

Answer 1

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

Answer 2

Convergence

Using the asymptotic approximation given in inequality $(10)$ 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 $p=3/2$.

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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 $$ \begin{align} \sum_{n=0}^\infty\frac{\prod\limits_{k=0}^n(2k+1)^2}{(2n+3)!} &=\sum_{n=0}^\infty\frac{\binom{2n}{n}(2n+1)}{4^n(2n+2)(2n+3)}\\ &=\arcsin(1)-1\\[6pt] &=\frac\pi2-1\tag7 \end{align} $$