I have the following series obtained via the Cauchy Product of the alternating harmonic series with itself
$\left( \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \right ) \cdot \left( \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \right ) = \sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{(-1)^k}{k} \cdot \frac{(-1)^{n-k}}{n-k}
=
\sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{(-1)^n}{k(n-k)}$
I wish to check the convergence of this Cauchy Product. I do not know how to handle double series. I do however know that I have to check if the following converges or diverges
$\sum_{n=1}^{\infty} c_n$
Where
$c_n = \sum_{k=1}^{n} \frac{(-1)^n}{k(n-k)}$
I just have no idea how to handle $c_n$. Can anyone please point out how to start?
P.S
The definition of the Cauchy Product I used is
$\left( \sum_{n=0}^{\infty} a_k \right) \cdot \left( \sum_{n=0}^{\infty} b_k \right) = \sum_{n=0}^{\infty} \sum_{k=0}^{n} a_k \cdot b_{n-k}$
Best Answer
Since
$$\sum_{n=1}^\infty \frac{(-1)^n}{n} = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{n+1},$$
we have, using the Cauchy product formula,
$$\begin{align}\left(\sum_{n=1}^\infty \frac{(-1)^n}{n}\right)\left(\sum_{n=1}^\infty \frac{(-1)^n}{n}\right) &= \left(\sum_{n=0}^\infty \frac{(-1)^{n+1}}{n+1}\right)\left(\sum_{n=0}^\infty \frac{(-1)^{n+1}}{n+1}\right)\\&= \sum_{n=0}^\infty \sum_{k=0}^n\frac{(-1)^{k+1}}{k+1}\frac{(-1)^{n+1-k}}{n+1-k}\\ &= \sum_{n=0}^\infty (-1)^n\sum_{k=0}^n\frac{1}{(k+1)(n+1-k)}\\ &= \sum_{n=0}^\infty (-1)^n c_n\end{align},$$
where
$$c_n = \sum_{k=0}^n\frac{1}{(k+1)(n+1-k)} = \sum_{k=0}^n\frac{1}{n+2} \left(\frac{1}{k+1} + \frac{1}{n+1-k} \right) \\= \frac{1}{n+2} \left(\sum_{k=0}^n \frac{1}{k+1} + \sum_{k=0}^n \frac{1}{n+1-k} \right) = \frac{2}{n+2}\sum_{k=0}^n\frac{1}{k+1} = \frac{2H_{n+1}}{n+2}$$
It is not difficult to show using the representation of $c_n$ in terms of the harmonic sum $H_{n+1}$ that $c_n$ is decreasing. We also have $c_n \to 0$ as $n \to \infty$. This follows from the fact that $H_{n+1} \sim \log(n+1)$. Alternatively, we can apply the Stolz-Cesaro theorem to get
$$\lim_{n \to \infty} \frac{H_{n+1}}{n+2}= \lim_{n \to \infty}\frac{H_{n+2} - H_{n+1}}{n+3 - (n+2)} =\lim_{n \to \infty} \frac{1}{n+2} = 0 $$
Therefore, the Cauchy product series converges by the alternating series test.