How to find the right b_n for limit comparsion test

Question

so we had this question in our homework test:

Find $\inf \alpha$ for which the series $\sum_{n=1}^\infty \frac{(2n)!}{n!(\alpha n)^n}$ converges

So I used the ratio test, and got that the limit is $\frac{4}{\alpha e}$. And by the ratio test, when $\alpha \gt \frac{4}{e}$ the series converges.

But I wonder what happens when $\alpha = \frac{4}{e}$, even though it isn't part of the question. according to WolframAlpha the series diverges by the limit test, but I couldn't figure which $b_n$ to use.

I should note that Stirling approximation isn't on the syllabus, so I can't use that.

Thanks

Answer 1

HINT

By Stirling approximation

$$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$

we have

$$\frac{(2n)!}{n!(\frac{4n}{e})^n}\sim \frac{\sqrt{4 \pi n}\left(\frac{2n}{e}\right)^{2n}}{\sqrt{2 \pi n}\left(\frac{n}{e}\right)^n}\left(\frac{e}{4n}\right)^n=\frac{\sqrt2\,4^nn^{2n}e^{2n}}{4^nn^{2n}e^{2n}}=\sqrt 2$$

Answer 2

Ratio test $$ \frac{a_{n+1}}{a_n}=\frac{\frac{(2n+2)!}{(n+1)!(\frac{4(n+1)}{e})^{n+1}}}{\frac{(2n)!}{n!(\frac{4n}{e})^n}}=\frac{(2n+1)e}{2(n+1)\big(\frac{n+1}{n}\big)^n}=\left(1-\frac{1}{2n+2}\right)\exp\left(1-n\log\Big(1+\frac{1}{n}\Big)\right) \\ = \left(1-\frac{1}{2n+2}\right)\exp\left(\frac{1}{2n}+\mathcal O\Big(\frac{1}{n^2}\Big)\right)=1+\mathcal O\Big(\frac{1}{n^2}\Big). $$ This implies that $a_n\not\to 0$, and hence the series diverges.