prove $\lim_{n→∞}\left(\frac{∑^{n}_{k=1}k^{m}}{n^{m+1}}\right)=1/(m+1)$

Question

I'm having trouble proving the following equation:

$$\lim_{n→∞}\left(\frac{∑^{n}_{k=1}k^{m}}{n^{m+1}}\right)=\frac{1}{(m+1)}$$

A link to a proof would suffice. Thank you.

Answer 1

Hint: Regard the expression as a limit of a Riemann sum, and evaluate the (easy) corresponding integral.

Answer 2

Use Stolz-Cesaro: $$\lim_{n\to\infty}\left(\frac{\sum^n_{k=1}k^{m}}{n^{m+1}}\right)=\lim_{n\to\infty}\frac{(n+1)^m}{(n+1)^{m+1}-n^m}=\cdots$$