$$ \lim_{n \to \infty} \frac{1^m + 2^m + 3^m + ... + (2n-1)^m }{n^{m+1}} $$

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?


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}

  • $\begingroup$ The $\approx$ should be $\lim_{n \rightarrow \infty}$ $\endgroup$
    – jwg
    Nov 2 '16 at 14:36
  • 1
    $\begingroup$ @jwg I know. I wanted the reader to think a little. $\endgroup$ Nov 2 '16 at 14:42
  • $\begingroup$ can you give some more details into the solution? thanks! $\endgroup$
    – user139708
    Nov 24 '16 at 2:40

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