I need find the exact value or at least to find a condition on $p$ and $k$ for which the following sum is finite: $$\sum_{n=1}^\infty\sum_{m=\left\lfloor n^{1/k}\right\rfloor }^\infty m^{-p}$$ for $p>1,k>1$.
We know (thanks alot to Meet Taraviya) that for $k=1$ $$\sum_{n=1}^\infty\sum_{m=n}^\infty m^{-p}=\sum_{n=1}^\infty n\cdot n^{-p}=\sum_{n=1}^\infty n^{1-p}=\zeta(1-p)$$ where where $\zeta(p)$ is the Riemann zeta function thus it converges for $p>2$.
Are there any other solutions for different values of $k$?
Thanks for any help.
Similarly when $k \geq 1$, since the terms are positive, you can rearrange the sums :
$$ \begin{align} \sum_{n=1}^\infty \sum_{m=\lfloor n^{1/k}\rfloor}^\infty m^{-p} &= \sum_{m=1}^\infty \sum_{n=1}^{(m+1)^k-1} m^{-p} \\ &=\sum_{m=1}^\infty ((m+1)^k-1)m^{-p}. \end{align} $$
Since
$$((m+1)^k-1)m^{-p} \sim m^{k-p}$$
when $m$ tends to $+\infty$, by comparison between series with non-negative terms, it converges iff $k-p<-1$, that is iff $p>k+1$.