Convergence of the power series $\sum \left(\frac{n^n}{n!} x^n \right)$ Find the convergence radius of the serie
$$\sum \frac{n^n}{n!}x^n $$
and analyze the absolute convergence and/or uniform.
What I've done:


*

*It is easy to show that the radius of convergence of this series is $R=\frac{1}{e}$. Then, the series convergence absolutely and uniformly on the interval$\; \left(-\frac{1}{e}, \frac{1}{e} \right)$

*Analyzing the convergence at $x=\frac{1}{e}$ (see here), the serie does not comverges.

*Analyzing the convergence at $x=\frac{-1}{e},$ the Dirichlet test says that the serie converges, if I didnt do anything wrong, using the same factorial aproximation we've seen before
I have some questions about what I've done and the difference between absolute and uniform convergence.


*

*Since the uniform convergence talks about series of functions, I think that does not makes sence to "analyze the uniform convergence" st the interval extremes. Is that correct?But it does makes sence to talk about uniform convergence at $[-\frac{1}{e}, \frac{1}{e})$! I am confuse about this part.

*I think the final answer should be that the series converges absolutely and uniformly at the interval of convergence and converges at the interval of convergence and for $x=\frac{1}{e},$ if I didnt do anything wrong.
I hope my questions are clear. Thanks for your help!
 A: The convergence radius $r$ of $\sum a_nx^n$ is given by:
$$
r^{-1}=\limsup_{n\to\infty}|a_n|^{1/n}
$$
By Stirling approximation:
$$
e=\lim_{n\to\infty}\left(\frac{e^n}{e\sqrt{ n}}\right)^{1/n}\leq\limsup_{n\to\infty}\left(\frac{n^n}{n!}\right)^{1/n}\leq\lim_{n\to\infty}\left(\frac{e^n}{\sqrt{2\pi n}}\right)^{1/n}=e
$$
Therefore $r=1/e$. The series converges absolutely for $|x|<1/e$ and converges uniformly on compact subset of $]-1/e,1/e[$. Since the series converges, as the PO said, for $-1/e$, it also converges uniformly on any compact subset of $[-1/e,1/e[$ (just pick the larger $N$ of the two as in $n>N\Rightarrow |f_n(x)-f(x)|<\varepsilon, \forall x\in]-1/e,1/e[$ and in $x=-1/e$).
The series is not uniformly convergent on $[-1/e,1/e[$, for otherwise $|\sum_{k=n}^mf_k(x)|\leq\varepsilon$ for $f_k(x)= \frac{k^k}{k!}x^k,\forall n,m>N, \forall x\in [-1/e,1/e[$ (uniformly cauchy=uniformly convergent on $\mathbb R$); as $x\to 1/e$, since $f_k(x)$ are all continuous, $|\sum_{k=n}^mf_k(1/e)|\leq\varepsilon$. We arrive at the conclusion that the series converges for $x=1/e$, which is a contradiction. 
In my intuitive understanding, since the series blows up at $x=1/e$, it cannot have a uniform convergence as $x$ approaches $1/e$.
