Determine all real $x$ for which the series $\sum\limits_{k=1}^\infty\frac{k^k}{k!}x^k$ converges. 
Determine all real $x$ for which the following series converges:
  $$\sum_{k=1}^\infty\frac{k^k}{k!}x^k.$$
  You may use the fact that
  $$\lim_{k\to\infty}\frac{k!}{\sqrt{2\pi k}(k/e)^k}=1.$$

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I found that the radius of convergence is $\displaystyle \frac{1}{e}$.Next considered end points. When $\displaystyle x=-\frac{1}{e}$, the series converges by alternating series test. But I was fail to determine it for the case where $\displaystyle x=\frac{1}{e}$. I tried to use comparison test by using the given fact. I guess in this case the series diverges. Can anybody please give me a hint to complete the proof?
 A: Let $x=1/e$, and let $a_k=\frac{k^k}{k!}(1/e)^k$.  We want to show that $\sum a_k$ diverges.
Let $\frac{k!}{\sqrt{2\pi k}(k/e)^k}=c_k$. We are told that $\lim_{k\to\infty} c_k=1$.
With some algebra, we can see that 
$$a_k=\frac{k^k}{k!}(1/e)^k=\frac{1}{c_k}\cdot \frac{1}{\sqrt{2\pi k}}.$$
Note that the limit of $\frac{a_k}{1/\sqrt{k}}$ as $k\to\infty$ is a non-zero constant. Thus, by the Limit Comparison Test, $\sum_{k=1}^\infty a_k$ diverges.
Remark: We want to show (conditional) convergence at $x=-1/e$, using the alternating series test. So we want to show that the terms are decreasing in absolute value and have limit $0$. That the limit is $0$ follows from the Stirling Formula, but that is not enough to prove monotonicity. 
Look instead at the ratio $\frac{b_{k+1}}{b_k}$, where $b_n=\frac{n^n}{n!}(1/e)^n$. After some simplification, we find that
$$\frac{b_{k+1}}{b_k}=\frac{(k+1)^{k+1}}{k^k\cdot (k+1)}(1/e)=\frac{1}{e}\left(1+\frac{1}{k}\right)^k.$$
It is a standard fact that $\left(1+\frac{1}{k}\right)^k$ increases monotonically, and has limit $e$. It follows that $\frac{1}{e}\left(1+\frac{1}{k}\right)^k\lt 1$, and therefore $b_{k+1}\lt b_k$.
A: As a side note, I think it is worth mentioning that by the Lagrange inversion formula we have:
$$ \forall x\in\left(-\frac{1}{e},\frac{1}{e}\right)\qquad W(x) = \sum_{n\geq 1}\frac{(-n)^{n-1}}{n!} x^n $$
with $W$ being the Lambert $W$ function, i.e. the inverse function of $x e^x$ in a neighbourhood of the origin. Trivial manipulations then leads to:
$$ \forall x\in\left(-\frac{1}{e},\frac{1}{e}\right)\qquad \sum_{k\geq 1}\frac{k^k}{k!}\,x^k = \frac{-W(-x)}{1+W(-x)}.$$
The convergence in the endpoints may be studied through Stirling's inequality:
$$ \sqrt{2\pi k}\left(\frac{k}{e}\right)^k \exp\left(\frac{1}{12k+1}\right)\leq k!\leq \sqrt{2\pi k}\left(\frac{k}{e}\right)^k \exp\left(\frac{1}{12k}\right).$$
