In one of my screenshots I found the following equality :
$$\sum_{n=2}^\infty\frac{n+4}{n(n^2+3n+2)2^{n+2}}\left(n-\sum\limits_{k=2}^n\zeta(n)\right)=\frac{3\ln(A)}{2}+\frac{\gamma}8+\frac{19}{48}-\frac{19\ln(2)}{24}-\frac{\ln(\pi)}{4}$$
with no explanation of what $A$ is.
Is A a known number ? If not, what does it equal here ? (I do not have the proof of the above, nor can I prove it myself, and therefore I can't find $A$ myself)
Edit :
I just found the generalisation $$\sum_{n=2}^\infty\frac{n(1-x)+2}{n(n+1)(n+2)}x^{n+1}\left(n-\sum\limits_{k=2}^n\zeta(n)\right)$$ $$=\ln(\Gamma(2-x))+\frac{3-\log(2\pi)}{2}x+\left(\frac{\gamma}{2}-1\right)x^2+\frac{x^3}{6}-x\ln(\Gamma(2-x))-\ln(G(2-x))$$
That might help in finding $A$ (i didn't manage to do it myself), but also raises another question for me : what is the function $G$ ?
The $G(z)$ in the generalization is the Barnes G function and $A$ in the first expression is the Glaisher-Kinkelin constant. Numerically${}^{\color{blue}{[1]}}$,
$$A \approx 1.282427129100622636875342568869791727767688927325001192063740\ldots$$
One can obtain the first expression from the generalization by fixing $x$ to $1/2$ and using following value of $G(z)$ at $z = \frac32$ ${}^{\color{blue}{[2]}}$. $$G\left(\frac32\right) = \frac{\sqrt[24]{2 e^3 \pi^6}}{A^{3/2}}$$
Notes