Given these two curious formulas:
How may you show that? $$\sum_{n=1}^{\infty}\frac{(2n-1)\zeta(2n+1)}{2^{2n+1}}\left[2n\left(1-\frac{1}{2^{2n}}\right)-\frac{(2n)^2+1}{(2n)^2-1}\right]=\gamma\tag1$$
$$\sum_{n=1}^{\infty}\frac{(2n-1)\zeta(2n+1)}{2^{2n+1}}\left[-2n\left(1-\frac{1}{2^{2n}}\right)+\frac{6n+1}{2n-1}\right]=\ln2\tag2$$
$\gamma=0.577216...; Euler-constant$
Any good hints.