Divergence at the endpoints of the interval of convergence for $\sum_{k=0}^{\infty}\frac{k!}{k^k}x^k$ It is easy enough to see from the ratio test that the radius of the series is $e$, but I struggled to show divergence of the series at $\pm e$. I am wondering in particular if there is a way using the series definition of $e$. Also I did not expect this sequence to diverge to infinity so any intuition for why $e^kk!$ beats $k^k$ so handily is welcome. 
My way:
$$
L=\lim_{k\rightarrow \infty}\frac{k!}{k^k}e^k\Rightarrow
\log(L)=\lim_{k\rightarrow \infty}k+\log(k!)-k\log(k)
$$
And then using stirling's, $k!=O(e^{-k}k^{k}\sqrt{k})$,
$$
\log(L)=\lim_{k\rightarrow \infty}k+\log(e^{-k}k^{k}\sqrt{k})-k\log(k)\\
=\lim_{k\rightarrow \infty}k-k+k\log(k)+\frac{1}{2}\log(k)-k\log(k)=\lim_{k\rightarrow \infty}\frac{1}{2}\log(k)=\infty
$$
 A: Well, to answer the one question about $e^k*k!$  compared to $k^k$, here's an intuition:   Consider each product consists of $k$ terms. one of them, each term is $k$.  The other, each term is $e*i$,  where $i$ ranges from $1$ to $k$.   Now, from $k/e$  onwards, each term is bigger than $k$,  so we get in percentages, only $1/e$ terms are less than $k$, or about $36\%$.  You could try proceeding onwards to match the bigger terms to smaller terms, and show that you have some bigger terms left over.
A: A possibility, not using Stirling but not with the series definition of $e$:
You have for $x\geq 0$ that $\log(1+x)\leq x$. This show that for $k\geq 1$, we have $\displaystyle (1+\frac{1}{k})^k=\exp(k\log(1+1/k))\leq e$. Now put $\displaystyle u_k=(-1)^k \frac{k!}{k^k} e^k$, or $\displaystyle u_k= \frac{k!}{k^k} e^k$. and $v_k=|u_k|$. It is easy to show that $\displaystyle \frac{v_{k+1}}{v_k}=\frac{e}{(1+1/k)^k}\geq 1$. Hence $v_k$ is increasing, and does not go to $0$ as $k\to +\infty$, and the series $u_k$ is divergent.  
A: From the Stirling formula, when we are dealing with $k >> 1$, 
$$k! \approx \sqrt{2\pi k}\left(\frac{k}{e}\right)^k$$
You can easily get
$$\frac{k!}{k^k} \approx \frac{\sqrt{2\pi k}}{e^k}$$
hence
$$e^k k!\approx k^k\sqrt{2\pi k}$$
Now
$$\sqrt{2\pi k} k^k \geq k^k$$
And so
$$e^k k! \geq k^k$$
