TITLE: Computing a tricky limit
QUESTION [7 upvotes]: $$ \lim_{n \to \infty} \frac{1^m + 2^m + 3^m + ... + (2n-1)^m }{n^{m+1}} $$
I am kind of stuck since I cannot make it look into a form that would involve the integral of certain function. I know somehow it would be easy if we can compare this limit to a riemman sum. Any ideas?

REPLY [14 votes]: Observe
\begin{align}
\sum^{2n-1}_{k=1} \frac{k^m}{n^m n} = \sum^{2n-1}_{k=1} \left(\frac{k}{n} \right)^m\frac{1}{n} \approx \int^2_0 x^m\ dx.
\end{align}