# Inverse of partial sums of general harmonic series

I would like to understand better the scaling of the following summation as a function of $r$ and $p > 1$:

$$S_r(p) := \sum_{m=1}^{r} \left( \sum_{k=r-m+1}^{r} \left( \frac{k^q}{\sum_{k'=1}^{r} (k')^q} \right)^{1/p} \right)^{-1} \:, \:\: \frac{1}{p} + \frac{1}{q} = 1 \:.$$

In particular, I am interested in upper bounds on $S_r(p)$. Let us consider the two extremes of $p$. When $p$ approaches $1$, we have that $S_r(1) = r$. On the other hand, when $p$ approaches $+\infty$, we have that $S_r(\infty) = H_r = \Theta(\log{r})$, where $H_r$ is the $r$-th Harmonic number. I would like to understand how to analyze the interpolation of $S_r(p)$ between $r$ and $\log{r}$. The intermediate $p=2$ case is also tractable, and is given by $S_r(2) = H_{2r} \sqrt{\frac{2r(1+r)}{3(1+2r)}} = \Theta(\sqrt{r} \log{r} )$.

I did some numerical calculations here. In particular, in the notebook there is a picture of the curve for a few values of $r$.

• I think for large $r$ one can replace all summations with integrals. – Nemo Oct 25 '17 at 5:31
• When $p$ is not close to $1$ it seems $S_r(p)=\Theta(r^{1/p}\log{r} )$, and more precisely $S_r(p)\sim\frac1{(1+q)^{1/p}}r^{1/p}\log{r}$. – Nemo Oct 25 '17 at 5:57