TITLE: Minimize $a$, but ensure convergence
QUESTION [5 upvotes]: I just answered a Math SE question asking whether or not
$$\sum_{n=1}^\infty \frac{(2n)!}{n!^2}$$
was divergent. Of course, it is obviously divergent. But then I became curious, and started thinking about the convergence/divergence of the sum
$$\sum_{n=1}^\infty \frac{(2n)!}{n!^a}$$
In the first problem, I had shown that each term was equal to
$$\frac{n+1}{1}\cdot\frac{n+2}{2}\cdot...\cdot\frac{n+n}{n}$$
and thus did not ever sink lower than $1$. However, if we instead took
$$\sum_{n=1}^\infty \frac{(2n)!}{n!^4}$$
The sum seems to converge (at least, according to Wolfram Alpha).
I would like to find the smallest $a$ for which the sum converges (or the largest for which it diverges). My guess is that it converges for all $a \gt 2$, but I have no idea how to prove this guess.
Does anybody have any hints for me?
Thanks!

REPLY [2 votes]: It is well-known and not difficult to prove that
$$ \frac{(2n)!}{n!^2}=\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}} \tag{1}$$
hence the absolute convergence of the given series for any $a>2$ also follows from asymptotic comparison. It might be interesting to notice that for $a=3$ we have
$$ \sum_{n\geq 0}\binom{2n}{n}\frac{1}{n!} = e^2 I_0(2)\approx\frac{101}{6} \tag{2}$$
with $I_0$ being a modified Bessel function of the first kind, for instance.