# A sum refers to euler's constant

Show that : $$\sum\limits_{m=1}^{\infty }{\sum\limits_{n={{2}^{m-1}}}^{{{2}^{m}}-1}{\frac{m}{\left( 2n+1 \right)\left( 2n+2 \right)}}}=1-\gamma$$

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\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}