TITLE: A sum refers to euler's constant
QUESTION [6 upvotes]: Show that :
$$\sum\limits_{m=1}^{\infty }{\sum\limits_{n={{2}^{m-1}}}^{{{2}^{m}}-1}{\frac{m}{\left( 2n+1 \right)\left( 2n+2 \right)}}}=1-\gamma $$

REPLY [7 votes]: $$
\begin{align}
&\sum_{m=1}^\infty\sum_{n=2^{m-1}}^{2^m-1}\frac{m}{\left(2n+1\right)\left(2n+2 \right)}\\
&=\sum_{m=1}^\infty\sum_{n=2^{m-1}}^{2^m-1}m\left(\frac1{2n+1}-\frac1{2n+2}\right)\\
&=\lim_{N\to\infty}\sum_{m=1}^N\left(\sum_{n=2^m}^{2^{m+1}-1}\frac{m}{n+1}-\sum_{n=2^{m-1}}^{2^m-1}\frac{m}{n+1}\right)\\
&=\lim_{N\to\infty}\left(\sum_{m=2}^{N+1}\sum_{n=2^{m-1}}^{2^m-1}\frac{\color{#C00000}{m}-\color{#00A000}{1}}{n+1}-\sum_{m=1}^N\sum_{n=2^{m-1}}^{2^m-1}\frac{\color{#0000FF}{m}}{n+1}\right)\\
&=\lim_{N\to\infty}\left(\color{#C00000}{\sum_{n=2^N}^{2^{N+1}-1}\frac{N+1}{n+1}}-\color{#0000FF}{\frac12}-\color{#00A000}{\sum_{n=2}^{2^{N+1}-1}\frac1{n+1}}\right)\\
&=\lim_{N\to\infty}\left((N+1)\left(\log(2)+O\left(2^{-N}\right)\right)+1-\sum_{n=1}^{2^{N+1}}\frac1n\right)\\
&=\lim_{N\to\infty}\left((N+1)\left(\log(2)+O\left(2^{-N}\right)\right)+1-\left(\gamma+\log\left(2^{N+1}\right)+O\left(2^{-N}\right)\right)\right)\\[6pt]
&=1-\gamma
\end{align}
$$