Is there a nice closed form for $\displaystyle \sum_{\substack{m\geq 2\\ n\geq 2}}\frac1{m^n-1}$ ?
It's easy to prove that $\displaystyle \sum_{\substack{m\geq 2\\ n\geq 2}}\frac1{m^n} = 1$ by summing first over $n$ (geometric sum) and then over $m$ (telescopic series).
It can be proven that $ \sum_{q \in Q} \frac1{q-1}=1 \quad \text{where } Q=\{m^n, n\geq 2, m\geq 2\}$. A natural extension is to compute $\displaystyle \sum_{m\geq 2, n\geq 2}\frac1{m^n-1}$.
Here's what I've noticed: $$\begin{align} \sum_{\substack{m\geq 2\\ n\geq 2}}\frac1{m^n-1}&=\sum_{\substack{m\geq 2\\ n\geq 2}}\frac{1/m^n}{1-1/m^n}=\sum_{\substack{m\geq 2\\ n\geq 2}}\sum_{k\geq 1}\frac1{m^{nk}}\\ &=\sum_{m\geq 2}\sum_{k\geq 1}\frac{1}{m^k(m^k-1)} = \sum_{m\geq 2}\sum_{k\geq 1}\left(\frac 1{m^k-1}-\frac 1{m^k}\right) \end{align}$$ but that looks like a dead-end.
According to Mathematica, an approximation of the limit is $1.130396449$.