Studying a series for convergence at the boundary points I came across this series.
$$\sum_{n=1}^\infty \frac{n!}{n^n}x^n$$
I was able to calculate its radius of convergence. If my calculations are OK, it is the number $e$. Is that correct?
Then I started wondering if the series is convergent or divergent for $x=\pm e$.
But I think I don't have any means to determine that. Is there any known (standard undergraduate calculus) theorem or theory which I can use for determining that? And also, if the series is convergent for $x=\pm e$, how do I calculate its sum?
If there's no general approach here, is there any trick which can be applied in this particular case?
 A: Note the limit $\frac{{n!}^{\frac 1n}}{n}\to \frac 1e$.
The above limit by Cauchy- Hadamard theorem gives the radius of convergence of the series $\sum_{n=1}^\infty \frac{n!}{n^n}x^n$ as $e$.
At $x=\pm e$, $|\frac{n!}{n^n}e^n|\sim\sqrt {2\pi n}$ (By Sterling's approximation) and hence by comparison test, the given series does not converge at $x=\pm e$ (as for convergence of series $\sum y_n$, the necessary condition is $y_n\to 0$).
A: Start with $\left(1 + \frac{1}{n} \right)^n$ is a strictly increasing sequence with limit $\mathrm{e}$.  (In some undergraduate calculus courses, this limit is the definition of $\mathrm{e}$.)  If we don't have strict monotonicity given, see the answers at Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing .
Now let's look at $x = \mathrm{e}$ and see if the sum converges.  It doesn't and we will show that by showing that the terms do not go to zero as $n \rightarrow \infty$.  \begin{align*}
\frac{(n+1)!}{(n+1)^{n+1}}\mathrm{e}^{n+1} - \frac{n!}{n^n}\mathrm{e}^n  
    &= \frac{(n+1)\cdot n!}{(n+1)(n+1)^{n}}\mathrm{e}\,\mathrm{e}^n - \frac{n!}{n^n}\mathrm{e}^n    \\
    &= \frac{n!}{n^n (1+1/n)^{n}}\mathrm{e}\,\mathrm{e}^n - \frac{n!}{n^n}\mathrm{e}^n    \\
    &= \frac{n!}{n^n}\mathrm{e}^n \left( \frac{1}{(1+1/n)^{n}}\mathrm{e} - 1 \right)
\end{align*}
Since $(1+1/n)^n$ is monotonically increasing to $\mathrm{e}$, the fraction $\frac{\mathrm{e}}{(1+1/n)^n}$ is greater than $1$ for all $n \geq 1$.  So the difference we study is positive, so the sequence of terms is strictly monotonically increasing.  When $n = 1$, the term is $\mathrm{e}$ and all subsequent terms are greater than this.  Since the terms in the series do not approach $0$ as $n \rightarrow \infty$, the series diverges.
Likewise,for $x = -\mathrm{e}$, the terms do not go to zero so the series diverges.
