Conditionally converges $\sum_{k=0}^\infty a_n$, $\sum_{k=0}^\infty b_n$ and also their Cauchy product

Question

Edited: I post a new post here that is somewhat related to this question.

It proved that:

If $\sum_{k=0}^\infty a_n$ and $\sum_{k=0}^\infty b_n$ converges conditionally (not absolutely), then their Cauchy product may converges or diverges by oscillation but never diverges to properly (to $\pm\infty$).

(See here for the proof that it won't diverges to $\infty$.)

As I was wondering, if $\sum_{k=0}^\infty a_n$ and $\sum_{k=0}^\infty b_n$ converges conditionally, can their Cauchy product converges conditionally?

Also, may it converges absolutely?
I believe it won't, by trying $(a_n)_{n=0}^\infty=(b_n)_{n=0}^\infty=(-1)^{n+1}\div(n+1)^3$ as an example that requiring Cauchy product to converge absolutely, one of the two series should converges absolutely (is it right?). Therefore, Cauchy product of two conditionally convergent series should not converges absolutely (how can I prove it?).

When I tried to prove the Cauchy product may converges conditionally, I encounter some problems.

If I let $(a_n)_{n=0}^\infty=(b_n)_{n=0}^\infty=\frac{(-1)^{n+1}}{(n+1)}$, I think I can prove, easily, that the Cauchy product doesn't converges absolutely.

My conjecture that the Cauchy product can converges conditionally is proved by answer by RRL, so from here on, my works can be ignored.

However, I need to prove $\vert c_{n}\vert\ge\vert c_{n+1}\vert$, that is, $$\vert c_{n}\vert-\vert c_{n+1}\vert\ge0$$ $$\sum_{k=0}^n\frac{1}{(k+1)(n+1-k)}-\sum_{k=0}^{n+1}\frac{1}{(k+1)(n+2-k)}\ge0$$

It seems to be true as I plugged it into Desmos. But perhaps trying to prove that $$-\frac{1}{(n+1)^3}-\frac{1}{(k+1)(n+1-k)}+\frac{1}{(k+1)(n+2-k)}\ge0$$ for all $k=0,...,n$ won't work?

I tried to just use some basic algebra method to prove, but I realised I can't, after making it very complicated and ugly.

Does answering my question here helps?

Any help will be appreciated. Thank you!

Answer 1

Your second example is the one that shows that the Cauchy product of conditionally convergent series may converge (only) conditionally. For $a_n = b_n = \frac{(-1)^{n+1}}{(n+1)^3}$ the series $\sum a_n, \sum b_n$ converge absolutely.

With $\sum a_n = \sum b_n = \sum_{n \geqslant 0} \frac{(-1)^{n+1}}{n+1}$, the Cauchy product is

$$\tag{*} \sum_{n=1}^\infty(-1)^n \sum_{k=1}^n \frac{1}{k(n+1-k)} = \sum_{n=1}^\infty\frac{(-1)^n(2)H_n}{n+1},$$

where the harmonic sum $H_n =\sum_{k=1}^n\frac{1}{k}$ appears.

To obtain the RHS of (*) note that,

$$\sum_{k=1}^n\frac{1}{k(n+1-k)} = \sum_{k=1}^n\frac{1}{n+1} \left(\frac{1}{k} + \frac{1}{n+1-k} \right) = \frac{1}{n+1} \left(\sum_{k=1}^n \frac{1}{k} + \sum_{k=1}^n \frac{1}{n+1-k} \right) \\ = \frac{2}{n+1}\sum_{k=1}^n\frac{1}{k} $$

Since $\frac{H_n}{n+1}$ decreases monotonically to $0$, (*) converges by the alternating series test.

However, the Cauchy product (*) is not absolutely convergent since

$$\frac{H_n}{n+1} \sim \frac{\gamma + \log n}{n+1} \quad \text{as}\,\, n \to \infty,$$

where $\gamma$ is the Euler-Mascheroni constat and we have divergence by the limit comparison test.

Examples of conditionally convergent series whose Cauchy product is absolutely convergent are given here.