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The $H_{n,r}$ generalized harmonic number is defined as: $$H_{n,r} = \sum_{k=1}^{n} \frac{1}{k^r}$$

I'm interested in the growth of $H_{n,r}$ as a function of $n$, for a fixed $r\in[0,1]$.


For $r>1$, $H_{n,r}=O(1)$ (as a function of $n$). For $r=1$, $H_{n,1}=O(\log n)$. For $r=0$, $H_{n,0}=n$.

How does $H_{n,r}$ grow for intermediate values of $r$?

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  • $\begingroup$ Hint: If $0<r<1$ then $H_{n,r}-\int_1^nt^{-r}\,dr$ is bounded. $\endgroup$ – David C. Ullrich Dec 20 '15 at 16:54
  • $\begingroup$ The following MSE link may prove useful reading. $\endgroup$ – Marko Riedel Dec 20 '15 at 21:02
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The Euler-Maclaurin Sum Formula is tailor-made for this kind of application. For $r\ne1$, $$ \sum_{k=1}^nk^{-r}=\zeta(r)+\frac1{1-r}n^{1-r}+\frac12n^{-r}-\frac{r}{12}n^{-r-1}+O\left(n^{-r-2}\right) $$ Note that for $r$ near $1$, we have $\zeta(r)=\frac1{r-1}+\gamma+O\left(r-1\right)$. Therefore, $$ \begin{align} \lim_{r\to1}\left(\zeta(r)+\frac1{1-r}n^{1-r}\right) &=\lim_{r\to1}\left(\frac{n^{1-r}-1}{1-r}+\gamma+O\left(r-1\right)\right)\\[3pt] &=\log(n)+\gamma \end{align} $$ This gives us the standard expansion for the Harmonic series: $$ \sum_{k=1}^n\frac1k=\log(n)+\gamma+\frac1{2n}-\frac1{12n^2}+O\left(\frac1{n^3}\right) $$

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  • $\begingroup$ I guess you mean to use $n$ on the RHS? $\endgroup$ – R B Dec 20 '15 at 16:55
  • $\begingroup$ @RB: yes. Fixed $\endgroup$ – robjohn Dec 20 '15 at 16:56

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