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Interested by this post, running cases, I have been able to observe that, for any value of $a$ (positive integer, rational, irrational, complex)
$$\frac{\sum_{i=1}^n i^a}{n^{a+1}}= \frac{H_n^{(-a)}}{(n+1)^{a+1}} =\frac{1}{a+1}+\frac{1}{2 n}+\frac{a}{12 n^2}+O\left(\frac{1}{n^3}\right)$$ I wonder how this could be directly obtained since all my attempts have been totally unsuccessful.

All ideas and suggestions will be very welcome.

Claude Leibovici
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  • Isn'it readily obtained from Faulhaber's formula? https://en.wikipedia.org/wiki/Faulhaber%27s_formula – René Gy May 12 '17 at 08:27
  • @RenéGy. I forgot to precise that this is for any value of $a$ (integer, rational or irrational). I shall edit. – Claude Leibovici May 12 '17 at 08:32
  • @ClaudeLeibovici though you probably already know this, probably worth noting that $$\sum_{i=1}^n i^a = \frac{1}{n^{-1-a}} - a\int_1^n\frac{\lfloor x\rfloor}{x^{1-a}}dx = \frac{1}{n^{-1-a}}+\sum_{k=1}^{n-1}k^{a+1}-\sum_{k=1}^{n-1}k\left(k+1\right)^a$$ – Brevan Ellefsen Oct 11 '17 at 03:13
  • @ClaudeLeibovici It may be possible to extract some information from that integral by generalizing the method in this post, which handles the case $a=-1$ (i.e. the harmonic numbers) – Brevan Ellefsen Oct 11 '17 at 03:22

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