# Evaluate $\sum \sum 1/n^k$

I wanted to evaluate the sum:

$$\sum_{n \ge 2} \left(\zeta(n) – 1\right)$$

I rewrote this as:

$$\sum_{n\ge 2} \sum_{k\ge 2} \frac{1}{n^k}$$

I tried exploiting the symmetry but that didn’t seem to help. I know from numerical calculation that the answer is 1.

#### Solutions Collecting From Web of "Evaluate $\sum \sum 1/n^k$"

As Shalop points out, the inner sum is the geometric series $\frac{1}{n^2}+\frac{1}{n^3}+\cdots=\frac{1/n^2}{1-1/n}=\frac{1}{n(n-1)}$ which admits the partial fraction decomposition $\frac{1}{n-1}-\frac{1}{n}$. Thus your sum simplifies to the telescoping series $(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\cdots$ which you should know how to do.