Asymptotic of $ \sum_{r=1}^{k} \frac1{r^{3/2}}$ - Generalized Harmonic Number

Question

What is the asymptotic approximation of the following generalized Harmonic number as $k \to \infty$ ?

$$H(1.5,k) = \sum_{r=1}^{r=k} \frac{1}{r^{1.5}}$$

(there is a similar question posted on MSE but there is a wrong comment bellow which refers the reader to wolfram alpha, the comment is wrong since the notation used on wolfram alpha is different from what the questioner used).

Answer 1

Hint. You may express you sum in terms of the Hurwitz zeta function: $$ \sum_1^k\frac{1}{r^{3/2}}=\zeta(3/2)-\zeta(3/2,k+1) $$ and the asymptotic behavior, as $k \to \infty$, is given by

$$ \sum_1^k\frac{1}{r^{3/2}}= \zeta(3/2)-\frac{2}{\sqrt{k+1}}-\frac12 \frac1{(k+1)^{3/2}}+\mathcal{O}\left( \frac1{k^{5/2}}\right) $$

(see the link above). You would obtain the same result with the Euler-Maclaurin formula.

Answer 2

There is an asymptotic Euler-Maclaurin expansion for generalized harmonic numbers:

$$\zeta(s)\sim H_{N,s}+N^{-s}\sum_{k\,\ge-1}\frac{B_{k+1}}{(k+1)!}\frac{s^{(k)}}{N^k} \quad \textrm{as }~N\to\infty$$

The $B_{k+1}$s are the Bernoulli numbers with $B_1:=+\frac{1}{2}$ and the $s^{(k)}=s(s+1)\cdots(s+(k-1))$ are rising factorials. One may of course isolate the $H_{N,s}$ term for an asymptotic with it instead.

Note $H_{N,s}:=\displaystyle\sum_{k=1}^N k^{-s}$ and $s\in\Bbb C\setminus\{1\}$. Source: e.g. the first equation here.