How does one show that $\sum_{k=2}^{\infty}{(-1)^k\over k+1}\cdot{ \lceil \log_2(k) \rceil}=1-2\gamma?$

Question

Consider

$$\sum_{k=2}^{\infty}{(-1)^k\over k+1}\cdot{ \lceil \log_2(k) \rceil}=1-2\gamma\tag1$$

How does on show that $(1)$ converges to $1-2\gamma?$

Answer 1

The given series equals $$ \sum_{k\geq 2}\frac{(-1)^k}{k+1}+\sum_{k\geq 3}\frac{(-1)^k}{k+1}+\sum_{k\geq 7}\frac{(-1)^k}{k+1}+\ldots \tag{1}$$ or $$ \log(2)-\frac{1}{2}+\int_{0}^{1}\left[\sum_{h\geq 2}\sum_{k\geq 2^h-1}(-x)^k\right]\,dx \tag{2}$$ hence the claim boils down to proving Catalan's integral$^{(*)}$ $$ \gamma=\int_{0}^{1}\frac{1}{1+x}\sum_{n\geq 1}x^{2^n-1}\,dx \tag{3}$$ where the RHS of $(3)$ equals $$ \int_{0}^{1}\frac{1}{x}\sum_{m\geq 0}(-1)^m x^m \sum_{n\geq 1}x^{2^n}\,dx = \int_{0}^{1} \frac{1}{x}\sum_{m\geq 2} r(m) x^m\tag{4}$$ with $r(m)$ being the difference between the number of ways we may represent $m$ as $2^a+2b$ and the number of ways we may represent $m$ as $2^a+(2b+1)$ with $a\geq 1$ and $b\geq 0$.

$(*)$ The linked page shows a derivation, $(24)\to(29)$, based on Euler's series acceleration method.