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While grading some basic coursework on analysis, I read an argument, that a Cauchy product of two series that converge but not absolutely can never converge i.e. if $\sum a_n$, $\sum b_n$ converge but not absolutely, the series $\sum c_n$ with $$c_n= \sum_{k=0}^n a_{n-k}b_k$$ diverges.
Although we didn't have any theorem in the course stating something like this, it made me wonder if it was true.
As Davide Giraudo has said in the comments, we can find a counter example by using $a_k = b_k = (-1)^k/k$ for $k\geq 1$ and $a_0=b_0=0.$ In that case we compute $$c_n = \sum_{k=0}^n a_{n-k} b_k = \sum_{k=1}^{n-1} \frac{(-1)^{n-k}}{n-k} \frac{(-1)^k}{k} = (-1)^n \sum_{k=1}^{n-1} \frac{1}{k(n-k)}$$
$$ = \frac{(-1)^n}{n} \sum_{k=1}^{n-1} \frac{n-k+k}{k(n-k)} = \frac{(-1)^n}{n} \sum_{k=1}^{n-1} \left( \frac{1}{k} + \frac{1}{n-k}\right) = 2\frac{(-1)^n}{n} \sum_{k=1}^{n-1} \frac{1}{k}.$$
We show $\sum c_n$ converges by applying the Leibniz criterion. $c_n \to 0$ is clear, so we need only verify that $d_n = \frac{1}{n} \sum_{k=1}^{n-1}\frac{1}{k} $ is monotonically decreasing for sufficiently large $n.$
We compute $$d_{n+1}- d_n = \frac{1}{n+1} \sum_{k=1}^n \frac{1}{k} - \frac{1}{n} \sum_{k=1}^{n-1} \frac{1}{k}= \frac{1}{n(n+1)} - \frac{1}{n(n+1)}\sum_{k=1}^{n-1} \frac{1}{k}.$$
Since $\displaystyle \sum_{k=1}^{n-1} \frac{1}{k} \geq 1$ for all $n\geq 2$ so $d_n$ is indeed monotone.
