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 $.

Recall that, formally,
$$
\left(\sum_{n=1}^{\infty} a_n\right)\left(\sum_{n=1}^{\infty} b_n\right) = \sum_{n=1}^{\infty} c_{n+1},$$
where
$$
c_n = \sum_{k=1}^{n-1} a_k b_{n-k}.
$$
If the series $\sum c_{n+1}$ converges, then the above equality is actually true.  You seem to know how to show this, so I'll just demonstrate the formal aspect of the problem.
Let $a_n = b_n = \frac{(-1)^{n}}{n}$.  Then
$$
a_k b_{n-k} = \frac{(-1)^n}{k(n-k)} = \frac{(-1)^n}{n}\left(\frac{1}{k}+\frac{1}{n-k}\right),
$$
so that
$$
\begin{align*}
c_n &= \frac{(-1)^n}{n} \sum_{k=1}^{n-1} \left(\frac{1}{k}+\frac{1}{n-k}\right) \\
&= 2\frac{(-1)^n}{n} \sum_{k=1}^{n-1} \frac{1}{k}.
\end{align*}
$$
We therefore have
$$
2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1} \sum_{k=1}^{n} \frac{1}{k} = \left(\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\right)^2 = (-\log 2)^2 = (\log 2)^2.
$$
