# 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?

• $$\frac{1}{k(k-1)} = \frac{1}{k-1} - \frac 1k$$ – Omnomnomnom Nov 28 '16 at 15:23

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 :)

• +1 if you put the phrase "Telescoping Series" in your answer so that he can research more into them. – AlgorithmsX Nov 28 '16 at 15:34
• @AlgorithmsX Enjoy! ;) – tired Nov 28 '16 at 15:44
• Thank you. It is near impossible to look up concepts in math without a connection to some name. – AlgorithmsX Nov 28 '16 at 15:48
• @tired That should do it! +1 – Mark Viola Nov 28 '16 at 16:04
• @Dr.MV Thanks a lot (as always) – tired Nov 28 '16 at 16:10