Why does $\sum_{k=2}^\infty k \left( \sum_{j=2^k}^{2^{k+1}-1} \frac{(-1)^j}{j} \right )+{1\over2}-{1\over3} = \gamma$? How could one prove that
$$x = \sum_{k=2}^\infty k \left( \sum_{j=2^k}^{2^{k+1}-1} \frac{(-1)^j}{j} \right )$$
is such that $x+{1\over2}-{1\over3} = \gamma$ ?
I am having problems just calculating to see if it is correct, let alone proving it...

Start with equation $(15)$ in this answer:
$$
\begin{align}
\gamma
&=\sum_{k=1}^\infty AHT(2^k-1)\\
&=\sum_{k=1}^\infty\sum_{j=0}^\infty\frac{(-1)^j}{2^k+j}\\
&=\color{#C00000}{\sum_{k=1}^\infty\sum_{j=2^k}^\infty\frac{(-1)^j}j}\\
&=\sum_{k=1}^\infty\sum_{i=k}^\infty\sum_{j=2^i}^{2^{i+1}-1}\frac{(-1)^j}j\\
&=\sum_{i=1}^\infty\sum_{k=1}^i\sum_{j=2^i}^{2^{i+1}-1}\frac{(-1)^j}j\\
&=\color{#C00000}{\sum_{i=1}^\infty i\sum_{j=2^i}^{2^{i+1}-1}\frac{(-1)^j}j}\\
&=\frac12-\frac13+\sum_{i=2}^\infty i\sum_{j=2^i}^{2^{i+1}-1}\frac{(-1)^j}j
\end{align}
$$
