# Divergence of $\sum_{k=1}^{\infty}\frac{k^k}{e^k}$

I'm trying to show the series $\sum_{k=1}^{\infty}\frac{k^k}{e^k}$ is divergent by the negation of the cauchy criterion. My thought was to break the sum into dyadic pieces that could be bounded from below to show the series divrges but I'm having trouble finding a lower bound. Any ideas, thanks?

• Apply ratio or root test. You'll se that it diverges – newzad Oct 22 '13 at 2:13
• The general term does not actually go to zero! That should do it, right? – RKD Oct 22 '13 at 2:14
• Why use the Cauchy criterion? The sequence of terms diverges, so of course the series does as well. – Brian M. Scott Oct 22 '13 at 2:14
• If $k\ge 3$, then $\frac{k^k}{e^k}\gt 1$. – André Nicolas Oct 22 '13 at 2:16

The limit of $k^k/e^k$ tends to infinity, therefore not to $0$, thus the sum diverges.
Let $a_k=\frac{k^k}{e^k}$.
$\log(a_k)=\log(k^k)-\log(e^k)=k(\log(k)-1)$
Thereby, $\log(a_k) \sim \infty$
We can lower bound each term past $k=8>e^2$ as $a_k>e^k$ since $\log(a_k)>1$.
Even a single $e^k$ term grows large exponentially quickly. Hope that helps.