Evaluate:: $ 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 +\cdots + \frac 1n\right) $ How to evaluate the series:
$$  2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 + \cdots + \frac 1n\right) $$
According to Mathematica, this converges to $ (\log 2)^2 $.

In this answer, it is shown that
$$
\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}H_n=\frac12\zeta(2)-\frac12\log(2)^2
$$
This sum is
$$
\begin{align}
2\sum_{n=1}^\infty\frac{(-1)^n}{n}H_{n-1}
&=2\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\left(\frac1n-H_n\right)\\
&=2\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}-2\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}H_n\\
&=\zeta(2)-\left(\zeta(2)-\log(2)^2\right)\\[6pt]
&=\log(2)^2
\end{align}
$$
