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

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