TITLE: What is the limit $\lim_{n\to\infty}\frac{1^n+2^n+3^n+\dots+n^n}{n^{n+1}}$?
QUESTION [10 upvotes]: 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!

REPLY [3 votes]: 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 $$\lim_{n\to\infty}\frac{1^n+2^n+3^n+\ldots+n^n}{n^{n+1}}=0.$$