How to solve $\lim\limits_{n \to \infty} \frac{\sum\limits_{k=1}^{n} k^m}{n^{m+1}}$ I'm trying to get the limit of $\lim\limits_{n \to \infty} \frac{\sum\limits_{k=1}^{n} k^m}{n^{m+1}}$
I know that $\sum\limits_{k=1}^{n} k^m$ is a polynom of $\deg(m +1)$ by the Euler-MacLaurin formula. And that I may apply L'hôspitals rule $m+1$ times.
So there will be some constant in the denominator and the numerator I guess? Maybe something like $(m+1)!$
But I can't find out what exactly yields the $m+1$ the derivitive of that sum. Has anybody a tipp on how to solve this?
Edit: I forgot the $+1$, sry
 A: Hint:
$$\frac{1}{n}\sum_{k=1}^n\left(  \frac{k}{n} \right)^m \approx  \int_{0}^{1}x^m\text{d}x $$
The above is true due to Riemann sum & how we define the integral. Therefore: $$ \lim_{n \to  \infty }   \frac{1}{n} \sum_{k=1}^{n} \left(  \frac{k}{n} \right)^m= \int_{0}^{1}x^m\text{d}x= \frac{1}{m+1} $$
A: Applying Stolz–Cesàro theorem, where $a_n=\sum\limits_{k=1}^n k^m$ and $b_n=n^{m+1}$ gives
$$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\\
\frac{(n+1)^m}{(n+1)^{m+1}-n^{m+1}}=\\
\frac{(n+1)^m}{(n+1-n)\left((n+1)^{m}+(n+1)^{m-1}n+(n+1)^{m-2}n^2+...+(n+1)n^{m-1}+n^m\right)}=\\
\frac{(n+1)^m}{(n+1)^{m}+(n+1)^{m-1}n+(n+1)^{m-2}n^2+...+(n+1)^2n^{m-2}+(n+1)n^{m-1}+n^m}=\\
\frac{1}{1+\frac{n}{n+1}+\frac{n^2}{(n+1)^2}+...+\frac{n^{m-2}}{(n+1)^{m-2}}+\frac{n^{m-1}}{(n+1)^{m-1}}+\frac{n^{m}}{(n+1)^{m}}}\to\frac{1}{m+1}, n\to\infty$$
Thus
$$\frac{a_n}{b_n}\to\frac{1}{m+1}, n\to\infty$$
A: Let $s(m,n) = \sum_{k=1}^n k^m$, then as you write, $s(n,m)$ is a polynomial in $n$ of degree $m+1$. Thus,
$$
\frac{s(m,n)}{n^m} = \Theta(n) \to \infty
$$
as $n \to \infty$.

A much more interesting question, of course, would we to explore the asymptotics of
$$
\frac{s(m,n)}{n^{m+1}} = \Theta(1).
$$
I would start at writing out $s(n,1), s(n,2), \ldots$ and contemplating the sequence of leading coefficients.
A: You can get more than the limit using generalized harmonic numbers
$$ \sum\limits_{k=1}^{n} k^m=H_n^{(-m)}$$
Using asymptotics
$$n^{-m} \sum\limits_{k=1}^{n} k^m=n^{-m}\,H_n^{(-m)}=n^{-m} \left(n^m \left(\frac{n}{m+1}+\frac{1}{2}+\frac{m}{12
   n}+O\left(\left(\frac{1}{n}\right)^3\right)\right)+\zeta (-m)\right)$$ Expand and get
$$\frac{n}{m+1}+\frac{m}{12 n}+\frac{1}{2}\to \infty$$
