jmc

We want to find the sum $$\begin{array}{rl}& \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots \\ & +\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots \\ & +\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\cdots \\ & +\cdots .\end{array}$$The sum of the numbers in the $n$th column is an infinite geometric series, with first term $\frac{1}{(n+1{)}^2}$ and common ratio $\frac{1}{n+1},$ so the sum of its terms is $$\frac{\frac{1}{(n+1{)}^2}}{1-\frac{1}{n+1}}=\frac{1}{n(n+1)}=\frac{(n+1)-n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.$$Hence, the sum of the terms is $$\sum _{n=1}^{\infty }\left(\frac{1}{n}-\frac{1}{n+1}\right)=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots =\boxed{1}.$$