# Minimize $a$, but ensure convergence

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!

You'll need to understand the size of $$\frac{(2n+2)(2n+1)}{(n+1)^a}$$ as $n\rightarrow \infty.$
• Okay, I see. I will end up with $$\frac{2n(2n-1)}{n^a}$$ and so this converges for all $n \gt 2$. Thanks for the great hint! I will gladly accept your answer as soon as I am allowed to. :D – Frpzzd Jun 27 '17 at 20:53
• @Jason the series converges for $a>2$, while is divergent for $a\le2$ – Ixion Jun 27 '17 at 22:32
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.