Mathematical Explanation of Mathematica Summation ${\sum_{n=1}^{\infty}\frac{(2n-1)!}{(2n+2)!}\zeta(2n)}$

Solution 1:

Only a note.

WolframAlpha can calculate $\enspace\text{Sum[Zeta[2n]*(2n-1)!/(2n+1)!,{n,1,Infinity}]}$

$\displaystyle \left(=\sum\limits_{n=1}^\infty \frac{\zeta(2n)(2n-1)!}{(2n+1)!}\right)\enspace$ exactly

but can only approximate $\enspace\text{Sum[Zeta[2n]*(2n)!/(2n+2)!,{n,1,Infinity}]}$

$\displaystyle \left(=\sum\limits_{n=1}^\infty \frac{\zeta(2n)(2n)!}{(2n+2)!}\right)$.

This is indeed strange, because the difficulty is about the same. WolframAlpha has problems with slow converging series.

Maybe the term $\,\,\text{$\zeta(3)/(8\pi^2$)}\,\,$ is by chance, because the calculation inaccuracy seems to be very close to this term at some point of the calculations.

It's interesting, that WolframAlpha calculates

$\,\,\text{Sum[Zeta[2n](2n-1)!/(2n+3)!,{n,1,Infinity}]}\,\,$ $\displaystyle \left(=\sum\limits_{n=1}^\infty \frac{\zeta(2n)(2n-1)!}{(2n+3)!}\right)$

exactly including $\,\text{$\zeta(3)/(8\pi^2$)}$, here $\,\text{$9\zeta(3)/(72\pi^2$)}\,$ .