Compute $\lim_{n\to\infty}n^{-(n+\nu)}\sum_{k=1}^nk^k$ for $\nu\in\Bbb R$ 
Let $\nu\in\Bbb R$ and 
  $$S_{\nu}:=\lim_{n\to\infty}\frac{1}{n^{n+\nu}}\sum_{k=1}^nk^k$$
  What is the value of $S_{\nu}$?

Disclaimer: This is not a Homework question. Indeed, I thought at this problem when reading this question.
We have
$$ n^{-\nu}\ =\ n^{n-(n+\nu)}\ \leq \ \underbrace{n^{-(n+\nu)}\sum_{k=1}^nk^k}_{=: \ S_{\nu}^n}\ \leq \ n^{n+1-(n+\nu)} \ = \ n^{1-\nu} \qquad \forall n \in \Bbb N$$
and thus
$ S_{\nu}=\infty$ if $\nu<0$ and $S_{\nu}=0$ if $\nu >1$.
Now, for $\nu\in [0,1]$, I have to admit that I don't see how to compute $S_{\nu}$. It seems that $S_{\nu}\in [0,1]$ for $\nu \in [0,1]$ as shown by the following plot of the few first values of $S_{\nu}^{n}$:


As pointed out in a comment, we have 
\begin{align*}n^n\ \leq\ \sum_{k=1}^nk^k \ &\ \leq n^n+(n-1)^{n-1}+\sum_{k=1}^{n-2}k^k\leq n^n+(n-1)^{n-1}+\overbrace{(n-2)(n-2)^{n-2}}^{=(n-2)^{n-1}} \\ &\leq n^n+2(n-1)^{n-1}\leq n^n+2n^{n-1}=n^n\Big(1+\frac{2}{n}\Big) \qquad \forall n\in\Bbb N.
\end{align*}
So that
$$n^{-\nu}\leq S_{\nu}^n \leq n^{-\nu}\Big(1+\frac{2}{n}\Big)\qquad \forall n\in\Bbb N.$$
Letting $n\to \infty$, we get
$$S_{\nu}=\begin{cases}\infty & \text{if } \nu<0,\\ 1 & \text{if }\nu=0, \\ 0 & \text{if } \nu >0.\end{cases}$$
