<s>[INST] The Cauchy product of $\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}$ with itself. In Apostol's Mathematical Analysis there is the following exercise.  Show that the Cauchy product of 
$$\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}$$
with itself is the series
$$2\sum_{n=1}^\infty{(-1)^{n+1}\over n+1}\left(1+\frac12+\cdots+\frac1n\right).$$
This doesn't seem to be true though. Writing out the terms,
$$\left(\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}\right)\left(\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}\right)=\frac11-\left(\frac11\frac12+\frac12\frac11\right)+\left(\frac11\frac13+\frac12\frac12+\frac13\frac11\right)+\cdots$$
and
$$2\sum_{n=1}^\infty{(-1)^{n+1}\over n+1}\sum_{k=1}^n\frac1k=2\left(-\frac12+\frac13\left(1+\frac12\right)-\frac14\left(1+\frac12+\frac13\right)+\cdots\right).$$
I can see that in the first sum, all terms of the form $\frac1n\frac1m$ with $m\ne n$ are doubled, but the terms $\frac1{n^2}$ don't seem to be accounted for. It would be nice if the sum of the missing terms was $0$, but in fact what we seems to have is
$$\begin{align}
\left(\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}\right)\left(\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}\right) & =2\sum_{n=1}^\infty{(-1)^{n+1}\over n+1}\sum_{k=1}^n\frac1k+\sum_{n=1}^\infty\frac1{n^2}\\
& =2\sum_{n=1}^\infty{(-1)^{n+1}\over n+1}\sum_{k=1}^n\frac1k+{\pi^2\over 6}.
\end{align}$$
Any help is appreciated.
 [/INST] The Cauchy product will be 
$$\sum_{n = 0}^\infty \sum_{k = 0}^n \frac{(-1)^{k+1}}{k+1} \frac{(-1)^{n-k+1}}{n-k+1} = \sum_{n = 0}^\infty \sum_{k = 0}^n \frac{(-1)^n}{(k+1)(n-k+1)}.$$
Now
\begin{align}\sum_{k = 0}^n \frac{(-1)^n}{(k+1)(n-k+1)} &= (-1)^n \sum_{k = 0}^n \frac{1}{n+2}\left(\frac{1}{n-k+1} + \frac{1}{k+1}\right)\\
&= \frac{(-1)^n}{n+2}\left\{\sum_{k = 0}^n \frac{1}{n-k+1} + \sum_{k = 0}^n \frac{1}{k+1}\right\}\\
&= \frac{(-1)^n}{n+2}\left\{\sum_{j = 0}^n \frac{1}{j+1} + \sum_{k = 0}^n \frac{1}{k+1}\right\} \quad (j = n - k)\\
&= \frac{2(-1)^n}{n+2}\sum_{k = 0}^n \frac{1}{k+1}\\
\end{align}
So the product is reduced to 
$$2\sum_{n = 0}^\infty \frac{(-1)^n}{n+2}\sum_{k = 0}^n \frac{1}{k+1} = 2\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n+1}\left(1 + \frac{1}{2} + \cdots + \frac{1}{n}\right).$$
</s>