TITLE: Assymptotics of the generalized harmonic number $H_{n,r}$ for $r < 1$
QUESTION [12 upvotes]: 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 $r$?

REPLY [11 votes]: 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:
$$
\sum_{k=1}^n\frac1k=\log(n)+\gamma+\frac1{2n}-\frac1{12n^2}+O\left(\frac1{n^3}\right)
$$