Is $\sum\limits_{n = 1}^\infty (-1)^n\frac{H_n}{n}=\frac{6\ln^2(2)-\pi^2}{12}$ a valid identity?
Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$