## real analysis – Evaluating \$sum _{k=1}^{infty }frac{H_{2k}}{k^3:4^k}binom{2k}{k}\$.

As the solutions here, I tried similar ways.

I tried using:
$$sum _{k=1}^{infty }frac{x^{2k}}{k:4^k}binom{2k}{k}=-2ln left(1+sqrt{1-x^2}right)+2ln left(2right)$$
$$-2sum _{k=1}^{infty }frac{1}{k:4^k}binom{2k}{k}int _0^1x^{2k-1}ln left(1-xright):dx=4int _0^1frac{ln left(1-xright)ln left(1+sqrt{1-x^2}right)}{x}:dx$$
$$-4ln left(2right)int _0^1frac{ln left(1-xright)}{x}:dx$$
$$sum _{k=1}^{infty }frac{H_{2k}}{k^2:4^k}binom{2k}{k}=4int _0^1frac{ln left(1-xright)ln left(1+sqrt{1-x^2}right)}{x}:dx+4ln left(2right)zeta left(2right)$$

But I can’t get the right manipulations so that I end up with:
$$sum _{k=1}^{infty }frac{H_{2k}}{k^3:4^k}binom{2k}{k}$$

I think I might have another way to go, I’ll edit as soon as i advance on it.