# What values of $a$ make the series $\displaystyle \sum_{n=0}^{\infty}\frac{n!2^n}{e^{an}n^n}$ convergent?

I would like to find what values of $a$ make the following series convergent

$$\sum_{n=0}^{\infty}\frac{n!2^n}{e^{\large an}n^n}$$

I started applying $n$-root but i don't know how to solve

$$\lim _{n\to +\infty}\frac{\sqrt[n]{n!}}{n}$$

If there are other ways to solve the problem i would be glad to know them too!
Thank you very much!

I do not want to use Stirling's formula

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Try applying Stirligs formula to $n!$ – Jas Ter Feb 25 '13 at 19:51

You can indeed avoid Stirling.

The ratio test will give you the answer for every $a\neq \log 2-1$, since $$\lim \frac{a_{n+1}}{a_n}=\frac{2}{e^{a+1}}.$$ The only difficulty here is $\lim n\log (n/(n+1))=\lim n\log(1-1/(n+1))=e^{-1}$ which follows from the equivalent $\log(1-u)\sim -u$ as $u$ approaches $0$.

Now for $a=\log 2 -1$, you get $$\sum \frac{e^n n!}{n^n}.$$

Note that $$\log n!=\sum_{k=1}^n \log k\geq \int_1^n\log x dx =n\log n-n+1>n\log n-n$$ so $$n!\geq \frac{n^n}{e^n}.$$ Hence your general term does not converge to $0$.

So the series diverges in this case.

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Sorry but i get $\lim \frac{a_{n+1}}{a_n}=\frac{2}{e^{a}}$ – user63534 Feb 25 '13 at 22:39
@user63534 I am afraid it is because you said that $\left(\frac{n}{n+1}\right)^n$ converges to $1$ while it converges to $e^{-1}$. Or something like that... – 1015 Feb 25 '13 at 22:42
You are totally correct again – user63534 Feb 25 '13 at 22:42
@user63534 I notice you have asked seven questions without accepting any answer. Since you are new here, maybe you simply don't know that you can accept an answer if it pleases you. But you don't have to. It's up to you. People here tend to consider that is is preferable to do so more often than never. – 1015 Feb 26 '13 at 2:15

Since $n!$ is about $(n/e)^n$, each term is about $(\frac{2n}{e^{a+1}n})^n =(\frac{2}{e^{a+1}})^n$ and this converges if $e^{a+1}>2$ and diverges if <2.

If this =2, the sum diverges by incorporating the $\sqrt n$ term in n!.

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