Is there a closed form expression for $$\sum_{n=2}^{\infty} (1 - \zeta(n)^{-1})$$
I know this sum converges, as
converges by comparison with $\zeta(2)$. And $$\begin{align} &\zeta(n) - 1 \geq 1-\zeta(n)^{-1}\\ &\zeta(n) + \zeta(n)^{-1} \geq 2\\ &\zeta(n) + \frac{1}{\zeta(n)} \geq 2 \end{align}$$
which is always true for any $\zeta(n)$, due to AM-GM inequality. Thus, $$1 = \sum_{n=2}^{\infty} (\zeta(n) - 1) \geq \sum_{n=2}^{\infty} (1 - \zeta(n)^{-1})$$
proving convergence.
Other evaluations of summations using the zeta function have been solved by using various representations of the zeta function. For example, consider the integral representation of $\zeta(n)$ $$\frac{1}{\Gamma(n)}\int_{0}^{\infty} \frac{x^{n-1}}{e^x - 1}dx = \frac{1}{(n-1)!}\int_{0}^{\infty} \frac{x^{n-1}}{e^x - 1}dx$$
Thus, $$\sum_{n=2}^{\infty} (1 - \zeta(n)^{-1)} = \sum_{n=2}^{\infty} \frac{\zeta(n)-1}{\zeta(n)} = \sum_{n=2}^{\infty} [1 - \frac{(n-1)!}{\int_{0}^{\infty} \frac{x^{n-1}}{e^x - 1}dx}]$$
The reciprocal of the integral makes the sum tricky to evaluate. Using other representations of the zeta function results in the same problem. Are there any methods to get a closed form for this summation?
Using the Dirichlet series for the reciprocal zeta function (suggested by Gary), we have
$$\sum_{n=2}^{\infty} (1 - \zeta(n)^{-1}) = \sum_{n=2}^{\infty} (1 - \prod_{p \text{ prime}} (1 - p^{-n})) = \sum_{n=2}^{\infty} (1 - \sum_{m=1}^{\infty}\frac{\mu(m)}{m^n}) = \sum_{n=2}^{\infty} \sum_{m=2}^{\infty}\frac{\mu(m)}{m^n}$$
$$=\sum_{m=2}^{\infty}\frac{\mu(m)}{m(m-1)} = -\sum_{m=1}^{\infty} \frac{\mu(m+1)}{m}$$
Computations suggest this value converges somewhere around $0.705211$.
Thanks.
