Calculate $\sum\limits_{i=2}^\infty\sum\limits_{j=2}^\infty \frac{1}{j^i}$ Calculate $\sum\limits_{i=2}^\infty\sum\limits_{j=2}^\infty \frac{1}{j^i}$. I am trying to figure out how to calculate this. I know it must be $\lim_{k\rightarrow\infty}\sum\limits_{i=2}^k\sum\limits_{j=2}^k\frac{1}{j^i}$ but I am not sure how to do this?
 A: Hint (as first proposed by achille hui in the comments above): $$\sum_{i=2}^\infty j^{-i}=\frac{1}{j(j-1)} \; , \;\;j>1$$
EDIT:
The above is a geometric series $\sum_{i=0}^\infty r^i=\frac{1}{1-r}$ with $r=\frac{1}{j}$ where you have to subtract the first two terms (since $i$ runs from $2$ and not $0$).
A: $$
\begin{align}
\sum_{i=2}^\infty\sum_{j=2}^\infty\frac1{j^i}
&=\sum_{j=2}^\infty\sum_{i=2}^\infty\frac1{j^i}\tag{1}\\
&=\sum_{j=2}^\infty\frac{\frac1{j^2}}{1-\frac1j}\tag{2}\\
&=\sum_{j=2}^\infty\frac1{j(j-1)}\tag{3}\\
&=\sum_{j=2}^\infty\left(\frac1{j-1}-\frac1j\right)\tag{4}\\
&=\lim_{n\to\infty}\sum_{j=2}^n\left(\frac1{j-1}-\frac1j\right)\tag{5}\\
&=\lim_{n\to\infty}\left(1-\frac1n\right)\tag{6}\\[12pt]
&=1\tag{7}
\end{align}
$$
Explanation:
$(1)$: since all terms are positive, apply Tonelli's Theorem
$(2)$: use the Formula for the Sum of a Geometric Series
$(3)$: simplify the summand
$(4)$: apply Partial Fractions
$(5)$: apply the definition of an Infinite Series
$(6)$: compute the Telescoping Sum
$(7)$: evaluate the limit
A: Notice:


*

*$$\sum_{n=a}^{\infty}\frac{b}{n^c}=b\zeta(c,a)\space\text{ when }b=0\vee\Re(c)>1$$

*$$\sum_{n=a}^{\infty}\frac{b}{c^n}=\frac{bc^{1-a}}{c-1}\space\text{ when }|c|>1$$


So, to solve your question:
$$\sum_{n=2}^{\infty}\sum_{m=2}^{\infty}\frac{1}{m^n}=\left[\sum_{n=2}^{\infty}\left[\sum_{m=2}^{\infty}\frac{1}{m^n}\right]\right]=\left[\sum_{n=2}^{\infty}\left[\zeta(n)-1\right]\right]=1$$
