<s>[INST] 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 $.
 [/INST] This is a special case of a more general result derived here.
$$S = \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n+1} \sum_{k=1}^n \dfrac1k$$
Recall that $\dfrac1k = \displaystyle \int_0^1 x^{k-1} dx$ and $\dfrac1{n+1} = \displaystyle \int_0^1 y^n dy$.
Now use the following fact.
$$\sum_{k=0}^{\infty} \int_0^1 (-z)^k dz = \lim_{n \to \infty} \int_0^1 \dfrac{1 - (-z)^n}{1+z} dz$$
The sequence of  functions $f_n(z) = \dfrac{1 - (-z)^n}{1+z}$ is dominated by the function $g(z) = \dfrac2{1+z}$ in the interval $[0,1]$, which is integrable. Hence, we can swap the limit and the integral to get that
$$\lim_{n \to \infty} \int_0^1 \dfrac{1 - (-z)^n}{1+z} dz = \int_0^1 \dfrac{dz}{1+z}$$
Hence,
$$S = \sum_{n=1}^{\infty} (-1)^{n+1} \int_0^1 y^n dy \left(\sum_{k=1}^n \int_0^1 x^{k-1} dx  \right) = \sum_{n=1}^{\infty} (-1)^{n+1} \int_0^1 y^n dy \left(\int_0^1 \dfrac{1-x^n}{1-x} dx  \right)$$
Hence,
$$S = \int_0^1 \int_0^1 \dfrac{\dfrac{y}{1+y} - \dfrac{xy}{1+xy}}{1-x} dy dx = \int_0^1 \int_0^1 \dfrac{y+xy^2-xy-xy^2}{(1+y)(1+xy)(1-x)} dx dy\\
=\int_0^1 \int_0^1 \dfrac{y}{(1+y)(1+xy)} dx dy = \int_0^1 \dfrac{\log(1+y)}{1+y} dy = \left. \dfrac{\log^2(1+y)}2 \right \vert_0^1 = \dfrac{\log^2(2)}2$$
The sum you are interested in is $2S$ and hence the answer is $\log^2(2)$.
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