We have $$A=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\right)=\ln(2)$$ $$B=\left(1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\cdots\right)$$ Here in $A$ $+-+-$, in $B$ $++--$, so $$B-A=\frac{1}{2}\left(\psi^{(0)}\left(\frac{3}{4}\right)-\psi^{(0)}\left(\frac{1}{2}\right)\right)$$ and $$AB=\ln(2)\left(\frac{1}{2}\left(\psi^{(0)}\left(\frac{3}{4}\right)-\psi^{(0)}\left(\frac{1}{2}\right)\right)+\ln(2)\right)\approx\frac{\pi}{4}$$ Here $\psi^{(0)}(z)$ - digamma function.
How and where can I calculate this constant?
$$B=\sum_{k\geq 0}\left[\frac{1}{4k+1}+\frac{1}{4k+2}-\frac{1}{4k+3}-\frac{1}{4k+4}\right]=\int_{0}^{1}\sum_{k\geq 0}x^{4k}(1+x-x^2-x^3)\,dx $$ $$ B = \int_{0}^{1}\frac{1+x-x^2-x^3}{1-x^4}\,dx = \int_{0}^{1}\frac{1+x}{1+x^2}\,dx = \color{red}{\frac{\pi}{4}+\frac{\ln 2}{2}}.$$