With $\mu(n)$ the Möbius function, I experimented with the following function:
$$f(s)=\sum_{n=1}^\infty \frac{\mu(n)-\mu(n+1)}{n^s}$$
and found that:
$$f(1)= 1+\sum_{m=2}^\infty \left(1-\frac{1}{\zeta(m)}\right)$$
which is equal to Niven's constant.
Numerical evidence suggests that $f(s)$ is rapidly converging for $\Re(s) > 0$ and I wonder whether any other closed forms or expressions in series with $\zeta$ do exist for $s\ne 1$.
Thanks.
By summation by parts, $$ f(s) = 1-\sum_{n\geq 1}\mu(n+1)\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right) \tag{1}$$ and working backwards: $$ 1-\frac{1}{\zeta(m)} = -\sum_{n\geq 2}\frac{\mu(n)}{n^m} = -\sum_{n\geq 1}\frac{\mu(n+1)}{(n+1)^m} \tag{2}$$ so: $$ \sum_{m\geq 2}\left(1-\frac{1}{\zeta(m)}\right) = -\sum_{n\geq 1}\frac{\mu(n+1)}{n(n+1)}\tag{3}$$ re-proving your identity. If $s=2$, $$\frac{1}{n^2}-\frac{1}{(n+1)^2} = \frac{2n+1}{n^2(n+1)^2} = \sum_{m\geq 2}\frac{m}{(n+1)^{m+1}} \tag{4}$$ hence: $$ f(2) = 1+\sum_{m\geq 2}m\left(1-\frac{1}{\zeta(m+1)}\right)\tag{5} $$ $$ f(3) = 1+\sum_{m\geq 2}\frac{m(m+1)}{2}\left(1-\frac{1}{\zeta(m+2)}\right)\tag{6}$$ and so on.