It can be easily proven that $$\sum_{s=2}^{\infty}(\zeta(s)-1)=1$$ I thought of evaluating a very similar looking sum: $$\sum_{s=2}^{\infty}(\zeta(s)^2-1)$$
- Mathematica cannot solve this, but gives an approximate value of $2.4653$.
- This converges by the comparison test. Here is what I did: the formula from This article gives $$\sum_{s=2}^{\infty}(\zeta(s)^2-1)=\sum_{m=1}^{\infty}\left(\frac{d(m)}{m-1}-1\right)$$ where $d(m)$ is the number of divisors of $m$. This equals $$\lim_{n\to\infty}\sum_{m=1}^{n}\frac{d(m)}{m-1}-n$$ But now I am stuck. What can be done now? Any help would be appreciated.