Partial sum of $p$-series

Question

I need to find the following sum for $p>1$: $$\sum_{n=1}^\infty\sum_{m=n}^\infty m^{-p}$$ My intuition says it is supposed to converge but I cannot find a proof for this. The formula can also be expressed as a: $$\sum_{n=1}^\infty( \zeta(p) -\sum_{m=1}^{n-1} m^{-p})$$ where $\zeta(p)$ is the Riemann zeta function (which is a positive constant for any $p>1$ as far as I know). So if you can help with evaluation of $\sum\limits_{m=1}^{n-1} m^{-p}$ this can also be very helpful.

EDIT: Thanks for the answers! Another question is $$\sum_{n=1}^\infty\sum_{m=\left\lfloor n^{1/k}\right\rfloor }^\infty m^{-p}$$ for some $k$.

Thanks a lot for any help.

Answer 1

Expanding the series:- $$1^{-p}+2^{-p}+3^{-p}+4^{-p}...$$ $$\space\space\space\space\space\space\space+2^{-p}+3^{-p}+4^{-p}...$$ $$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space+3^{-p}+4^{-p}...$$

The number $m$ occurs $m$ times. Hence Series simplifies to $$\sum_{n=1}^\infty n\cdot n^{-p}=\sum_{n=1}^\infty n^{1-p}=\zeta(p-1)$$ This converges for $p>2$ and diverges for $p\le 2$

Answer 2

Note that for $p>1$, $\sum_{m=1}^\infty m^{-p}$ is convergent. Moreover $$0<\sum_{n=2}^\infty\sum_{m=n}^\infty m^{-p}\leq \sum_{n=2}^\infty\int_{x=n-1}^\infty \frac{1}{x^p}dx=\frac{1}{p-1}\sum_{n=2}^\infty\frac{1}{(n-1)^{p-1}}$$ which converges for $p>2$. What happens for $1<p\leq 2$? We have that $$\sum_{n=1}^\infty\sum_{m=n}^\infty m^{-p}\geq \sum_{n=1}^\infty\int_{x=n}^\infty \frac{1}{x^p}dx=\frac{1}{p-1}\sum_{n=1}^\infty\frac{1}{n^{p-1}}=+\infty.$$