What is the limit $\lim_{n\to\infty}\frac{1^n+2^n+3^n+\dots+n^n}{n^{n+1}}$? I want to determine the limit given below:
$$\lim_{n\to\infty}\frac{1^n+2^n+3^n+\ldots+n^n}{n^{n+1}}$$
I have tried to solve thise several times ,but with no results.I have tried using lema stolz cezaro and managed to find a general formula for the summation, but couldnt prove it
$$\frac{1}{n}\sum_{k=0}^{n-1}\left(1-\frac{k}{n}\right)^n$$
Edit: Thank you for your responses,but i want to ask you if it is possible to solve the problem by not using integrals :D. NOTEIm in the eleventh grade and i had been given this problem by my math tutor ,so, as a consequence i dont know how to interpretate the great majority of your answers.Thanks for your time ,gents!
 A: Note that $\sum_{j=1}^n j^n \le \int_{1}^{n+1} x^n$. 
(And recall that $((n+1)/n)^{n}$ has a finite limit.)
A: Here's another point of view. For each $n\in\mathbb N$ choose an integer $k(n)\in\{0,1,\ldots,n\}$ such that the following two properties are satisfied: $$\lim_{n\to\infty}\frac{k(n)}{n}=1\qquad\text{and}\qquad\lim_{n\to\infty}\left(\frac{k(n)}{n}\right)^n=0.\tag{$\ast$}$$ Some possible choices are: $$k(n)=\lfloor n-\log{n}\rfloor\qquad\text{or}\qquad k(n)=\lfloor n-\sqrt n\rfloor\qquad\text{or}\qquad k(n)=\lfloor n^{\frac{n}{n+1}}\rfloor,$$ where $\lfloor x\rfloor$ is the largest integer not exceeding $x$, i.e. the floor function. (That $(\ast)$ holds follows e.g. from $\lim_{n\to\infty}(1+\frac1n)^n=e$ in the first two cases and from $\lim_{n\to\infty}n^{\frac1n}=1$ in the third case.)
Given such a sequence $k(n)$, we have the following bounds: $$\begin{align}0&\leq\frac{1^n+2^n+\ldots+k(n)^n+(k(n)+1)^n+\ldots+n^n}{n^{n+1}}\\&\leq\frac{k(n)k(n)^n+(n-k(n))n^n}{n^{n+1}}\\&=\left(\frac{k(n)}{n}\right)^{n+1}+1-\frac{k(n)}{n}.\end{align}$$ This last sequence converges to zero by $(*)$, so our sequence is bounded between two sequences converging to zero. We may conclude that $$\lim_{n\to\infty}\frac{1^n+2^n+3^n+\ldots+n^n}{n^{n+1}}=0.$$
A: You only need to exploit the identity (proved here):
$$\sum_{k=0}^{K}\binom{n+k}{n}=\binom{n+K+1}{n+1}$$
and the trivial bound:
$$ m!\binom{n}{m} \leq n^m \leq m!\binom{n+m}{m}$$
to have:
$$ \forall m\in\mathbb{N},\quad \lim_{n\to +\infty}\frac{1^m+2^m+\ldots+n^m}{n^{m+1}}=\frac{1}{m+1}$$
from which it follows that your limit is zero. 
As an alternative approach, since over $[0,1]$ we have $1-x\leq e^{-x}$:

$$ \frac{1}{n}\sum_{k=0}^{n}\left(1-\frac{k}{n}\right)^n \leq \frac{1}{n}\sum_{k=0}^{n}e^{-k}\leq\frac{1}{n}\sum_{k=0}^{+\infty}e^{-k}=\frac{e}{n(e-1)}.$$

