Find limit $\lim_{n\to \infty}\frac{\sum_{k=1}^n k^n}{n^n}$. Now I want to find the limit 
 $$\lim_{n\to\infty}\frac{\sum\limits_{k=1}^n k^n}{n^n}.$$
I try to use the Stolz theorem as follows:
$$\lim_{n\to \infty}\frac{\sum_{k=1}^n k^n}{n^n}=
  \lim_{n\to \infty}\frac{\sum_{k=1}^{n+1} k^{n+1}-\sum_{k=1}^n k^n}{(n+1)^{n+1}-n^n}$$
$$=\lim_{n\to\infty}\frac{(n+1)^{n+1}+\sum_{k=1}^n(k^{n+1}-k^n)}{(n+1)^{n+1}-n^n}$$
$$=1+\lim_{n\to\infty}\frac{\sum_{k=1}^n(k^{n+1}-k^n)}{(n+1)^{n+1}}.$$
It seems to deal with the summation like this form:
$$\sum_{k=1}^{n}k^p,\text{with}\ p=n,n+1.$$
I have no way to deal this summation, any help and hint will welcome!
 A: Hint: For fixed $k,$ $$\frac{(n-k)^n}{n^n}=\left(1-\frac{k}n\right)^n\to e^{-k}$$

Details:
Note that for $k=0...,n-1$ $$\log\left(1-\frac{k}{n}\right)=-\frac{k}{n}-\frac{k^2}{2n^2}-\cdots\leq -\frac{k}{n}$$
So $$\left(1-\frac{k}{n}\right)^n \leq e^{-k}$$
Then apply the dominated convergence theorem by defining:
$$f_n(k)=\begin{cases}0 & k\geq n\\ \left(1-\frac{k}{n}\right)^n& 0\leq k<n\end{cases}$$
Then $|f_n(k)|\leq g(k)=e^{-k}$ and for any $k,$ $\lim_{k\to\infty}f_n(k)\to g(k).$
A: We have that
$$\frac{\sum\limits_{k=1}^n k^n}{n^n}=\sum_{k=1}^n \left( \frac k n \right)^n=\sum_{k=0}^{n-1} \left( \frac {n-k} n \right)^n=\sum_{k=0}^{n-1} \left( 1-\frac {k} n \right)^n$$
then, following the suggestion given by ComplexYetTrivial, let consider
$$a_n=\sum_{\substack{k=0 \\ k<n}}^{\infty} \left( 1-\frac {k} n \right)^n$$
which is strictly increasing and bounded indeed by AM-GM we have 


*

*$\sqrt[n+1]{\left(1-\frac{k}{n}\right)^n \cdot 1} \leq \frac{n
   \left(1   - \frac{k}{n}\right) + 1}{n+1} = 1 - \frac{k}{n+1} $


and 


*

*$\left( 1-\frac {k} n \right)^n=e^{n\log \left( 1-\frac {k} n
   \right)}=e^{n \left( -\frac {k} {n}-\frac {k^2} {2n^2}-\frac {k^3}
   {3n^3}-\ldots \right)}\le e^{-k}$


therefore by the monotone convergence theorem $a_n$ has finite limit and since for $k$ fixed


*

*$\left( 1-\frac {k} n \right)^n\to e^{-k}$


we have
$$\lim_{n\to \infty}\frac{\sum\limits_{k=1}^n k^n}{n^n}=\lim_{n\to \infty} \sum_{k=0}^{n-1} \left( 1-\frac {k} n \right)^n=\lim_{n\to \infty} \sum_{\substack{k=0 \\ k<n}}^{\infty} \left( 1-\frac {k} n \right)^n=\sum_{k=0}^{\infty}e^{-k}=\frac{e}{e-1}$$
