$\sum_{n=0}^\infty \frac{(1+\frac{1}{n})^n\cdot n!}{n^n}x^n$ prove if it converge or diverge for $x=-e$ $$\sum_{n=0}^\infty \frac{(1+\frac{1}{n})^n\cdot n!}{n^n}x^n$$
Hello, this power series has a radius of $e$, but I cannot conclude if it diverges or converges in $x = -e$, I didn't succeed writing it formally. I would to receive hints or a valid solution, thanks.
 A: You can prove that $a_n=\frac{(1+\frac{1}{n})^n\cdot n!}{n^n}e^n$ is a strict increasing sequence. So its limit is not zero and this implies your series is divergent!
$$\frac{a_{n+1}}{a_n}=\frac{(1+\frac{1}{n+1})^{n+1}\cdot (n+1)!}{{(n+1)}^{n+1}}e^{n+1}\frac{n^n}{e^n(1+\frac{1}{n})^n\cdot n!}
=\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^n}\frac{e}{(1+\frac{1}{n})^n}>1.$$
Here the sequence $\{(1+\frac{1}{n})^n\}$ is strict increasing tending to $e$. 
A: Note that
$$\tag 1\ln \left (\frac{n!e^n}{n^n}\right ) = \sum_{k=1}^{n} \ln k  + n - n\ln n.$$
Now $\ln x$ is increasing, and this implies $\sum_{k=1}^{n} \ln k \ge \int_1^n \ln x\, dx.$ That integral equals $n\ln n -n +1.$ Thus the expression in $(1) $ is at least $1.$ Therefore $\dfrac{n!e^n}{n^n}$ is at least $e,$ which implies your series diverges.
A: Use Stirling's approximation as noted by @Robert Z
$$n! \sim \sqrt{2 \pi n} \frac{n^n}{e^n}$$
So
$$ \sum_{n=0}^\infty \frac{(1+\frac{1}{n})^n\cdot n!}{n^n}x^n \sim \sqrt{2 \pi}\sum_{n=0}^\infty f(n)$$
where $$f(n) = (1+\frac{1}{n})^n\cdot \sqrt{ n} \frac{1}{e^n}x^n$$
For $x = -e$, we have
$$f(n) = (-1)^n(1+\frac{1}{n})^n\cdot \sqrt{ n} $$
Consider
Let's work with $\vert f(n) \vert$
$$g(n) = \log \vert f(n) \vert = \underbrace{n \log(1+\frac{1}{n})}_{\rightarrow 1} + \underbrace{\frac{1}{2} \log n}_{\rightarrow \infty} \rightarrow +\infty$$
So $$\vert f(n) \vert \rightarrow + \infty \neq 0$$
Hence, the series diverges.
A: $f_m(x)
=\sum_{n=1}^{m} \frac{(1+\frac{1}{n})^n\cdot n!}{n^n}x^n
$
(The usual)
$\begin{array}\\
(1+1/n)^n
&=\exp(n\ln(1+1/n))\\
&=\exp(n(1/n-1/(2n^2)+O(1/n^3)))\\
&=\exp(1-1/(2n)+O(1/n^2))\\
&=e\exp(-1/(2n)+O(1/n^2))\\
&=e(1-1/(2n)+O(1/n^2))\\
\end{array}
$
so,
since
$n! \approx cn^{n+1/2}e^{-n}
$,
$\begin{array}\\
f_m(x)
&=\sum_{n=1}^{m} \dfrac{(1+\frac{1}{n})^n\cdot n!}{n^n}x^n\\
&=\sum_{n=1}^{m} \dfrac{(e(1-1/(2n)+O(1/n^2)))\cdot n!}{n^n}x^n\\
&=\sum_{n=1}^{m} \dfrac{e n!}{n^n}x^n
-\sum_{n=1}^{m} \dfrac{e\cdot n!}{2n\cdot n^n}x^n
+O(\sum_{n=1}^{m} \dfrac{ n!}{n^2n^n}x^n)\\
&=\sum_{n=1}^{m} \dfrac{e cn^{n+1/2}e^{-n}}{n^n}x^n
-\sum_{n=1}^{m} \dfrac{e\cdot cn^{n+1/2}e^{-n}}{2n\cdot n^n}x^n
+O(\sum_{n=1}^{m} \dfrac{ cn^{n+1/2}e^{-n}}{n^2n^n}x^n)\\
&=ec\sum_{n=1}^{m} \sqrt{n}(x/e)^n
-(ec/2)\sum_{n=1}^{m} n^{-1/2}(x/e)^n
+O(\sum_{n=1}^{m} n^{-3/2}(x/e)^n)\\
\end{array}
$
At $x=e$,
the third series converges,
the second series diverges slowly
like $\sqrt{m}$,
and the first series diverges
like $m^{3/2}$.
So the overall sum diverges.
This holds even if
we take more terms
in the Stirling approximation,
since the first sum remains the same
and only the second sum is changed.
A: A necessary condition for a series $\sum _{k=0} ^{\infty} a_k $ to converge is that $\lim _{k \rightarrow \infty} a_k =0 $. You can prove this condition does not hold for $x=-e$, hence the power series diverges at $x=-e$. In fact, using Stirling approximation $a_k \approx e\sqrt{2\pi k} \left( \frac{x} {e} \right) ^k$ for $k \rightarrow \infty$. 
