$p^{th}$ power harmonic series We know $a_n=\sum_1^n 1/i$ is of o$(n)$ by 'Cesaro sum'. I was wondering whether such a conclusion can be made for $a_n=\sum_1^n \frac{1}{i^p}$ for $p \in (0,1)$ i.e. $a_n=$o($n^q)$ for suitably chosen $q<1$, because $q \geq 1$ does the job trivially . By Cauchy integral test if $q >1-p$ then the above holds. Can we say something stronger? Like $iff$ condition?

The function $x^{-p}$ is a positive decreasing function. For such functions, sums at evenly spaced points are well approximated by integrals. More precisely, $$ \int_1^n \frac{1}{x^p}dx < \sum_{i = 1}^n \frac{1}{i^p} < \int_1^n \frac{1}{x^p}dx + 1.$$ The $1$ appearing above is really the first term of the sum.

So in this case, we have that $$ \sum_{i \leq n} \frac{1}{i^p} \approx \frac{n^{1 - p}}{1-p},$$ and this is the correct asymptotic.

This works also for $p = 1$, except that the integral is then $\log n$. This is much stronger than the claim that it's merely bounded above by $n$.