looking for an argument to show the convergence in $\sum_n\frac{n!}{(en)^n}$ I am looking for an argument why $\sum_n{(n/e)^n\over n!}$ is convergent.
This series appeared after I was looking for a radius in $\sum_n \frac{n^n}{n!}x^n$ and found, that it is between $-1/e$ and $1/e$. Now I am looking for an argument , why the first one is convergent and do not know what to do.
Thank you
Sorry, it was a wrong formula-it is otherway around.I tried to change it.Thank you for your help
 A: For points at the ends of the convergence interval, the quotient and root test fail exactly because they are the end points. In some cases the refined quotient tests that are called Raabe, Kummer or Gauß test will provide a definite answer. But theory says that for any test, however refined, there will always be undecidable cases.

The Raabe test requires the consideration of 
$$n\left(\left|\frac{a_{n+1}}{a_n}\right|-1\right)=\frac{n}{e}\left((1+\tfrac1n)^n-e\right)$$
and this would need to have a bound smaller than $-1$.
A: Using Stirlings approximation, you can replace $n!/n^n$ by $e^{-n} \sqrt{2 \pi n}$.
This reduces $\sum\frac{n!}{n^n} x^n $ to $\sum e^{-n} \sqrt{ 2 \pi n}\, x^n $, where obviously the convergence is limited to $|x|< e$.
For $x= +e (1-\varepsilon)$ the limit for $\varepsilon \to 0$ goes to infinity, but the case of $x= -e (1-\varepsilon)$ gives us a set of (ugly) limits for $\varepsilon\to0$ that depends on the approximation order used for Stirlings approximation.
A: By Stirling, the general term of $$\sum_{n=1}^\infty\frac{(ne^{-1})^n}{n!}$$ is asymptotic to
$$\frac{(ne^{-1})^n}{\sqrt{2\pi n}(ne^{-1})^n}=\frac1{\sqrt{2\pi n}},$$ making the series diverge.
