Find $\sum_{k=1}^{\infty}\frac{1}{2^{k+1}-1}$ 
Calculate $$\sum_{k=1}^{\infty}\frac{1}{2^{k+1}-1}.$$ 

I used Wolfram|Alpha to compute it and got it to be approximately equal to $0.6$. How to compute it? Can someone give me a hint or a suggestion to do it? 
 A: It is a fast-convergent series, and a Lambert series, too, since
$$ S=\sum_{k\geq 1}\frac{1}{2^{k+1}-1}=-1+\sum_{k\geq 1}\frac{1}{2^k-1}=-1+\sum_{k\geq 1}\sum_{m\geq 1}\frac{1}{2^{mk}}=-1+\sum_{n\geq 1}\frac{d(n)}{2^n}$$
where $d(n)$ is the number of divisors of $n$. Since $d(n)\leq n$ (this is a very crude bound)
$$ S+1-\sum_{n=1}^{N}\frac{d(n)}{2^n}\leq \sum_{n>N}\frac{n}{2^n}=\frac{N+2}{2^N} $$
hence by choosing $N=30$ we get that 
$$ -1+\sum_{n=1}^{30}\frac{d(n)}{2^n} = \color{red}{0.6066951}49\ldots $$
is an extremely good approximation of $S$, with the correct red digits.

As suggested by Yves Daoust, another good strategy comes from noticing that $\frac{1}{2^k-1}$ is pretty close to $\frac{1}{2^k}$ if $k$ is large, hence we may perform a series acceleration in the following way:
$$ S=\sum_{k\geq 2}\frac{1}{2^k-1}=\frac{1}{2^2-1}+\sum_{k\geq 3}\frac{1}{2^k}+\sum_{k\geq 3}\frac{1}{2^k(2^k-1)}$$
turning $S$ into 
$$ S = \frac{7}{12}+\frac{1}{2^3(2^3-1)}+\sum_{k\geq 4}\frac{1}{4^k}+\sum_{k\geq 4}\frac{1}{4^k(2^k-1)} $$
or
$$ S = \frac{815}{1344}+\frac{1}{4^4(2^4-1)}+\sum_{k\geq 5}\frac{1}{8^k}+\sum_{k\geq 5}\frac{1}{8^k(2^k-1)} $$
so that $S$ equals $\frac{260927}{430080}$ plus $\sum_{k\geq 5}\frac{1}{8^k(2^k-1)}$. With just three iterations of this technique we get already $S=\color{red}{0.60669}41\ldots$, and and the fourth step we get $S\approx \frac{1391613}{2293760}=\color{red}{0.6066951}20\ldots$
In a compact form, this acceleration technique leads to:
$$ S = \sum_{k\geq 1}\left(\frac{1}{2^{(k^2-1)}(2^{k+1}-1)}+\frac{1}{2^{(k^2+k)}(2^k-1)}\right) $$
collapsing to:
$$\boxed{S=\frac{1}{4}+\sum_{k\geq 2}\frac{8^k+1}{(2^k-1)\,2^{k^2+k}}=0.6066951524152917637833\ldots} $$
with a significant convergence boost. Now the main term of the sum behaves like $2^{-k^2}$ instead of $2^{-k}$.
A: Maybe it is interesting to see that there is a “closed form” of this series in terms of the $q$-polygamma function. We have that $$S=\sum_{k\geq1}\frac{1}{2^{k+1}-1}=-1-\sum_{k\geq1}\frac{1}{1-2^{k}}
 $$ and recalling the definition of the $q$-polygamma function we have $$S=\color{red}{-1-\frac{\psi_{1/2}\left(1\right)+\log\left(1/2\right)}{\log\left(2\right)}}\approx0.60669515.$$
