Asymptotic behavior of divergent $p$-series

Question

I am intertested in the asymptotic scaling behavior of the divergent $p$-series $$ \sum_{k=1}^n \frac{1}{k^p} $$ for $0<p<1$, i.e., is there a closed-form sequence $a_n$ so that $$ \lim_{n \to \infty} \frac{\sum_{k=1}^n \frac{1}{k^p}}{a_n}=c $$ for some $0<c<\infty$? I am tempted to conjecture that $a_n=n^{1-p}$ (clearly works for $p=0$), but could not find a way to prove this.