How do I show that $E-\gamma=\lim_{j\to \infty}\sum_{n=1}^{j}n\left({1\over 2^n-1}-{1\over 2^n}+\cdots-{1\over 2^{n+1}-2}\right)$ Given the Erdos-Borwein's constant $E=\sum_{n\geq 1}\frac{1}{2^n-1}$ and the Euler-Mascheroni constant
$\gamma=0.5772156...=\sum_{n\geq 1}\left[\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right]$
how can I show that
$$E-\gamma=\lim_{n\to\infty}\left[1\left(1-{1\over 2}\right)+2\left({1\over 3}-{1\over 4}+{1\over 5}-{1\over 6}\right)+3\left({1\over 7}-{1\over 8}+\cdots+{1\over 13}-{1\over 14}\right)+\cdots+n\left({1\over 2^n-1}-\cdots-{1\over 2^{n+1}-2}\right)\right]$$
Any hints? 
The only series came to my mind is
$$\ln{2}=1-{1\over 2}+{1\over 3}-\cdots$$
Let take parts of the series
$$1+{2\over 3}+{3\over 7}+{4\over 14}+\cdots=2.74403...$$ not something I recognised of as a closed form.
 A: We have:
$$ E=\sum_{n\geq 1}\frac{1}{2^n-1}=\sum_{n\geq 1}(2^{-n}+2^{-2n}+\ldots)=\sum_{n\geq 1}\frac{d(n)}{2^n} $$
and:
$$ 1\left(1-\frac{1}{2}\right)+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+\ldots+n\left(\frac{1}{2^n-1}+\ldots-\frac{1}{2^{n+1}-2}\right) $$
equals:
$$ \sum_{k=1}^n k\left(A_{2^{k+1}-2}-A_{2^k-2}\right),\qquad A_k = \sum_{h=1}^{k}\frac{(-1)^{h+1}}{h} $$
where the alternating harmonic numbers $A_k$ fulfill
$$ A_{2k} = H_{2k}-H_{k} $$
and the sum of interest is so:
$$ \sum_{k=1}^{n}k\color{blue}{\left(H_{2^{k+1}-2}-H_{2^k-1}-H_{2^k-2}+H_{2^{k-1}-1}\right)}$$
The trick is now to recognize a telescopic structure in the blue term, then exploit a suitable reindexing and summation by parts. Obviously:
$$ \sum_{k\geq 1}\left(H_{2^k-1}-H_{2^k-2}\right) = \sum_{k\geq 1}\frac{1}{2^k-1}=\color{red}{E}\tag{1}$$
and:
$$ H_{n} = \color{red}{\gamma} + \log(n)+\frac{1}{2n}+O\left(\frac{1}{n^2}\right)\tag{2}$$
gives:
$$ \lim_{m\to +\infty}\left[(m+1)H_{2^m}-m H_{2^{m+1}}\right]=\color{red}{\gamma}, \tag{3}$$
completing the proof, since
$$ \sum_{k=1}^{n}k\left(\color{green}{H_{2^{k+1}-2}}\color{purple}{-H_{2^k-1}}\color{green}{-H_{2^k-2}}\color{purple}{+H_{2^{k-1}-1}}\right)= \color{green}{S^+(n)}-\color{purple}{S^{-}(n)}\tag{4}$$
where:
$$ \color{green}{S^+(n)} = \sum_{k=1}^{n}k\left(H_{2^{k+1}-2}-H_{2^k-2}\right) = n H_{2^{n+1}-2}-\sum_{k=1}^{n}H_{2^{k}-2} \tag{5}$$
and:
$$ \color{purple}{S^-(n)} = \sum_{k=1}^{n}k\left(H_{2^{k}-1}-H_{2^{k-1}-1}\right) = (n+1) H_{2^{n}-1}-\sum_{k=1}^{n}H_{2^{k}-1}.\tag{6}$$
