It is well known and easy to evaluate, that $\sum_{n=2}^{\infty}\left(\zeta (n)-1\right)=1$. I am trying to evaluate $$\sum_{n=2}^{\infty}\left(\frac{1}{\zeta (n)}-1\right)$$ The classical way gives me $$\sum_{n=2}^{\infty}\left(\frac{1}{\zeta (n)}-1\right)=\sum_{n=2}^{\infty}\frac{\mu(n)}{n\cdot(n-1)}=-1+\sum_{n=2}^{\infty}\frac{\mu(n+1)-\mu(n)}{n}=1+\sum_{n=2}^{\infty}\frac{\mu(n)}{n-1}$$ where $\mu$ is the Mobius function. But it dosn't help me at all.
WolframAlpha gives the result -0.705211 (exactly). EDIT The value -0.705211 is not exact, although WolframAlpha gives = and not ≈. A more accurate value is -0.705211140...
Q: How to evaluate $\sum_{n=2}^{\infty}\left(\frac{1}{\zeta (n)}-1\right)$?
Thank you very much.