# Cauchy-Product of non-absolutely convergent series

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.

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I'm probably missing something, but if we take $a_k:=\frac{(-1)^k}k=:b_k$ for $k\geq 1$ and $a_0=b_0=0$, we have $c_n=2\frac{(-1)^n}n\sum_{k=1}^{n-1}\frac 1k$, and this series is converging (but not absolutely). – Davide Giraudo Jun 10 '12 at 10:12
@Davide Giraudo I get $c_n = \sum_{k=0}^n a_{n-k} b_k = \sum_{k=1}^{n-1} \frac{(-1)^{n-k}}{n-k} \frac{(-1)^k}{k} = (-1)^n \sum_{k=1}^{n-1} \frac{1}{k(n-k)}.$ – Ragib Zaman Jun 10 '12 at 10:20
@RagibZaman: They're the same. – anon Jun 10 '12 at 10:26
@anon ahh I see. Thanks. – Ragib Zaman Jun 10 '12 at 10:28
ha ... you're right Davide. Taking the same series I also got to $c_n=2\frac{(-1)^n}n\sum_{k=1}^{n-1}\frac 1k$ but then didn't see that $\frac1n\sum_{k=1}^{n-1}\frac 1k$ is converging to zero. Thanks for helping out. – Peter Patzt Jun 10 '12 at 11:21

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 $d_n = \frac{1}{n} \sum_{k=1}^{n-1}$ 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.