# Assymptotics of the generalized harmonic number $H_{n,r}$ for $r < 1$

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 $$r$$?

• Hint: If $0<r<1$ then $H_{n,r}-\int_1^nt^{-r}\,dr$ is bounded. – David C. Ullrich Dec 20 '15 at 16:54
• The following MSE link may prove useful reading. – Marko Riedel Dec 20 '15 at 21:02

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: $$\sum_{k=1}^n\frac1k=\log(n)+\gamma+\frac1{2n}-\frac1{12n^2}+O\left(\frac1{n^3}\right)$$
• I guess you mean to use $n$ on the RHS? – R B Dec 20 '15 at 16:55