Evaluate: $$\sum_{n=1}^{\infty} {\left(\frac{(-1)^n}{n}\sum_{k=1}^n {\left(\frac{(-1)^k}{k}\right)}\right)}$$ It looks like the series for $\ln(2)$ 'embedded in itself', so my guess for the value is $\ln^2(2)$. Unfortunately, this is not correct, as an estimate for the sum is $1.06269$, much greater than $\ln^2(2)\approx 0.48045$.
Any ideas? Thanks!
Let $S$ be given by
$$S=\sum_{n=1}^\infty \frac{(-1)^n}{n}\sum_{k=1}^n \frac{(-1)^k}{k}\tag1$$
We can interchange the order of summation in $(1)$ and express $S$ as
$$S=\sum_{k=1}^\infty \frac{(-1)^k}{k}\sum_{n=k}^\infty \frac{(-1)^n}{n} \tag2$$
whereupon switching "dummy" summation indices in $(2)$ yields
$$\begin{align}S&=\sum_{n=1}^\infty \frac{(-1)^n}{n}\sum_{k=n}^\infty \frac{(-1)^k}{k}\\\\ &=\sum_{n=1}^\infty \frac{(-1)^n}{n}\left(\sum_{k=1}^\infty \frac{(-1)^k}{k}-\sum_{k=1}^{n-1} \frac{(-1)^k}{k}\right)\\\\ &=\log^2(2)-\sum_{n=1}^\infty \frac{(-1)^n}{n} \sum_{k=1}^{n-1} \frac{(-1)^k}{k}\\\\ &=\log^2(2)-S+\sum_{n=1}^\infty\frac{1}{n^2}\tag3 \end{align}$$
Using $\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$ in $(3)$ and solving for $S$ we find
$$\bbox[5px,border:2px solid #C0A000]{S=\frac12 \log^2(2)+\frac{\pi^2}{12}}\tag 4$$
NOTE:
The result in $(4)$ is not at all surprising given the symmetry. We can write in general
$$\begin{align} \left(\sum_{n=1}^\infty a_n \right)^2&=\sum_{n=1}^\infty a_n \sum_{k=1}^\infty a_k\\\\ &=2\sum_{n=1}^\infty a_n \sum_{k=1}^{n} a_k-\sum_{n=1}^\infty a_n^2 \end{align}$$
which becomes upon rearranging
$$\sum_{n=1}^\infty a_n \sum_{k=1}^{n} a_k=\frac12 \left(\sum_{n=1}^\infty a_n \right)^2+\frac12 \sum_{n=1}^\infty a_n^2$$
Finally, setting $a_n=\frac{(-1)^n}{n}$, we recover the result in $(4)$.
