Sum of series $\sum_{k=1}^\infty{-\frac{1}{2^k-1}}$ What is the sum of the series $\displaystyle\sum_{k=1}^\infty{\frac{-1}{2^k-1}}$?
Also, more generally, can we find $\displaystyle\sum_{k=1}^\infty{\frac{-1}{c^k-1}}$ for some $c$?

Ignoring the negative sign, you're looking for
$$ {1 \over 2^1 - 1} + {1 \over 2^2 - 1} + {1 \over 2^3 - 1} + \cdots $$
but we have
$$ {1 \over 2^{jk} - 1} = {1 \over 2^{jk}} + {1 \over 2^{2jk}} + {1 \over 2^{3jk}} + \cdots. $$
If we rewrite the first expression using the second one, then your sum is
$$ \left( {1 \over 2^1} + {1 \over 2^2} + {1 \over 2^3} + \cdots \right) + \left( {1 \over 2^2} + {1 \over 2^4} + {1 \over 2^6} + \cdots \right) + \left( {1 \over 2^3} + {1 \over 2^6} + {1 \over 2^9} + \cdots \right) + \cdots $$
and $1/2^r$ appears $\tau(r)$ times, where $\tau(r)$ is the number of divisors of $r$. Therefore your sum is
$$ \sum_{r \ge 1} \tau(r) 2^{-r} $$
and this should allow you to compute it to any desired level of numerical accuracy fairly quickly. For example,
$$ \sum_{r=1}^{40} \tau(r) 2^{-r} = {27602812537 \over 17179869184} = 1.606695152 $$
and the error here is less than $\sum_{r \ge 41} r 2^{-r} = 84/2^{41} < 4 \times 10^{-11}$ since $\tau(r) < r$ for all $r \ge 3$.
Of course this method works when $2$ is replaced by any constant $c > 1$.
