Showing $\sum_{n=1}^{\infty }(-1)^{n-1}\frac{(2n-2)!\zeta (2n)}{\pi ^{2n}}(1-\frac{1}{2^{2n}})(1+\frac{1}{2^{2n-1}})=\frac{\log(2)}{4}$.

Question

How do I show that show $$\frac{0!\zeta (2)}{\pi ^2}\left(1-\frac{1}{2^2}\right)\left(1+\frac{1}{2}\right)-\frac{2!\zeta (4)}{\pi ^4}\left(1-\frac{1}{2^4}\right)\left(1+\frac{1}{2^3}\right)+\dots-\dots=\frac{\log(2)}{4}?$$

Answer 1

Not an answer but kinda hint: $${\rm B}_{2n}=(-1)^{n-1}\frac{(2n)!\zeta(2n)}{2^{2n-1}\pi^{2n}}$$ So: $${\rm S}=\sum_{n=1}^{\infty}\frac{(2^{2n}-1)(2^{2n-1}+1){\rm B}_{2n}}{2^{2n}(2n)(2n-1)}$$