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 $c$?
 A: 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 $c > 1$.
A: Wolfram Alpha gives an explicit answer for c=2, approximately -1.6067, and an explicit mess for arbitrary c.
A: $$\sum_{k=1}^\infty  c^{-k}\left(1-\left(\frac{1}{c} \right)^k \right)^{-1} ;c>1 $$
$$\sum_{k=1}^\infty c^{-k}  +\sum_{k=1}^\infty c^{-2k} +\sum_{k=1}^\infty  c^{-3k} \cdots = \sum_{k=1}^\infty \sum_{n=1}^\infty c^{-n k} $$
In the double summation written above note that ($s=c^{-1}$) , $s^l$ appears as many times as there are solutions to  $nk=l$ where $n,k,l\in \mathbb{N}$. Let $f(l)$ be the number of solutions, then 
$$\sum_{k=1}^\infty \sum_{n=1}^\infty s^{n k} = \sum_{l=1}^\infty f(l) s^l ; s<1  $$
The coefficient $f(l)$ is studied in more detail in number theory as it appears in the square of the zeta function. $\zeta (s)^2 = \sum_{n=1}^\infty \frac{f(n)}{n^s}$
The $f(n)$ is called the divisor function. So one way will be to write the series as a power series with the coefficients $f(n)$. 
Other option is to use  Computer algebra, as Ross answered. Mathematica gives a closed form
$$\frac{-\text{Log}[-1+c]-\text{Log}\left[\frac{1}{c}\right]-\text{QPolyGamma}\left[0,1,\frac{1}{c}\right]}{\text{Log}[c]}$$
A: It is the Erdős-Borwein Constant. (modulo a minus sign)
