is $\sum_{m=2}^{\infty}{\left(\sum_{n=2}^{\infty}\frac{1}{n^m}\right)}$ divergent or convergent

Question

is $$\sum_{m=2}^{\infty}{\left(\sum_{n=2}^{\infty}\frac{1}{n^m}\right)}$$ divergent or convergent. If its divergent are there any lower bounds for n and/or m that would make it convergent. I know that it is divergent if n or m is less than 1 but I'm not sure for higher values

Answer 1

Reversing the order of summation, this is $$\sum_{n=2}^\infty\sum_{m=2}^\infty\frac1{n^m} =\sum_{n=2}^\infty\frac{1}{n(n-1)}=1$$ as this series telescopes.

Answer 2

Since for any $m\geq 2$ we have $\zeta(m)-1=\frac{1}{(m-1)!}\int_{0}^{+\infty}\frac{x^{m-1}}{e^x(e^x-1)}\,dx $ it is pretty straightforward that

$$ \sum_{m\geq 2}\left(\zeta(m)-1\right) = \int_{0}^{+\infty}\frac{dx}{e^x(e^x-1)}\sum_{m\geq 2}\frac{x^{m-1}}{(m-1)!}=\int_{0}^{+\infty}e^{-x}\,dx = 1.$$

Answer 3

The series is: $$\sum_{m=2}^{\infty} \left(\frac{1}{2^m}+\frac{1}{3^m}+\frac{1}{4^m}+\cdots\right)$$ Use the ratio test: $$\lim_\limits{m\to\infty} \frac{a_{m+1}}{a_m}=\lim_\limits{m\to\infty} \frac{\frac{1}{2^{m+1}}+\frac{1}{3^{m+1}}+\frac{1}{4^{m+1}}+\cdots}{\frac{1}{2^m}+\frac{1}{3^m}+\frac{1}{4^m}+\cdots}<1$$ The series converges.