Evaluate $\sum_{k=2}^{\infty}\left(\sum_{n=2}^{\infty} {1 \over k^n}\right)$ I am trying to evaluate
$$\sum_{k=2}^{\infty}\left(\sum_{n=2}^{\infty} {1 \over k^n}\right)$$
First observation: ${1 \over k^n} < 1$. So, in the limit, the sum of $p$ first terms of the inner sum is
$$\lim \limits_{p \to \infty}\left({1 \over k^2}{1 - {1 \over k^p} \over 1 - {1 \over k}}\right) =  {1 \over k^2}{1 \over 1 - {1 \over k}} = {1 \over k(k - 1)}$$
The outer sum now looks like this
$$\sum_{k=2}^{\infty}{1 \over k(k - 1)} = {1 \over 2} + {1 \over 6} + {1 \over 12} + {1 \over 20} + \dots$$
We know that ${1 \over k^2}$ converges, so the rewritten sum will converge as well. Do you have any ideas how to find the sum?

Hint: 
$$
\frac{1}{2}+
\color{blue}{\frac{1}{6}}+
\color{red}{\frac{1}{12}+}
\color{green}{\frac{1}{20}}+...=\frac{1}{2}+
\color{blue}{\frac{1}{2}-\frac{1}{3}}+
\color{red}{\frac{1}{3}-\frac{1}{4}}+
\color{\green}{\frac{1}{4}-\frac{1}{5}}+...
$$
which is an example for a telescoping series
Btw: You have proven that $\sum_{n\geq1}(\zeta(n)-1)=1$, congratulations :)
