Series whose Cauchy product is absolutely convergent - A general example

Question

Is there series that is divergent or conditionally convergent with absolutely convergent Cauchy product? Seems like there is a group of these examples!

Perhaps finding divergent series with absolutely convergent Cauchy product isn't that difficult (see here for an example), but perhaps finding conditionally convergent series with absolutely convergent Cauchy product is more complicated.

I've been studying the paper by Florian Cajori after RRL posted a comment in my post:
Conditionally converges $\sum_{k=0}^\infty a_n$, $\sum_{k=0}^\infty b_n$ and also their Cauchy product .

Perhaps the paper is a bit 'ancient' (100 years ago) (perhaps mathematicians 100 years ago had a style a little bit differently than modern day mathematicians). Perhaps mathematicians 100 years ago had a style a little bit differently than modern day mathematicians.

Or else, I think my mathematical maturity isn't enough, so I used lots of time to read his paper but still fails to understand all materials in the paper. (For example, I only realize what is meant by 'dropping parentheses' after I read it for an hour...)

The author of the paper gives two examples, the first example is about, there are conditionally convergent series such that their Cauchy product are absolutely convergent. That example, is digested by me after some works.

My works on this example, hope that there aren't typos or errors :). P.1 P.2 P.3 P.4 P.5

He gives 'the general method' later in the paper. That's what I'm stuggling.

Not to say that I can't understand why the constraints are set that way, I even don't know how to prove the general method (and the author can't write down his delicate proof because the margin is too small).

Following his words, I conjectured that:

If $\hspace{2ex}$$b,c\in\mathbb N, a=2c;\vert d\vert=\min\{\vert l\vert:b\equiv l\text{ }(\text{mod }a)\}, d\mid c$;

$\hspace{4ex}$$\forall(k=1,...,c),\gamma_k=1$; $\forall (k=c+1,...,a), > > \gamma_k=-1$;

$\hspace{4ex}$$\forall(k=1,...,b),\delta_k\in\{-1,1\}$;

$\hspace{4ex}$$\forall (n\ge m_1,k=1,...,a),s_{n,k}=\frac{\gamma_k} {an+\alpha_k},\text{ while }m_1\in\mathbb Z$

$\hspace{8ex}$is chosen s.t. $(s_{n,k})$ are well-defined;

$\hspace{4ex}$$\forall (n\ge m_2,k=1,...,b), t_{n,k}=\frac{\delta_k} {bn+\beta_k},\text{ while }m_2\in\mathbb Z$

$\hspace{8ex}$is chosen s.t. $(t_{n,k})$ are well-defined;

$\hspace{4ex}$$\forall(n\ge m_1,k=1,...,a), u_{an+k}=s_{n,k}$;

$\hspace{4ex}$$\forall (n\ge m_2,k=1,...,b), v_{bn+k}= t_{n,k}$;

$\hspace{4ex}$$m\ge \max\{am_1+1,bm_2+1\}$ is an arbitrary integer.

Then $\sum_{n=m}^\infty u_n$ converges conditionally and Cauchy product of $\bigg(\sum_{n=m}^\infty u_n,\sum_{n=m}^\infty v_n\bigg)$,

$\hspace{4ex}$i.e. $\sum_{n=m}^\infty c_n=\sum_{n=m}^\infty \bigg( \sum_{k=m}^n u_k v_{n+m-k} \bigg)$ converges absolutely.

Here's what I've tried:

I hope to break (It is called 'dropping paraentheses', as mentioned in the paper) the series $\sum_n^\infty \Big(c_{nab+1}+c_{nab+2}+...+c_{nab+(ab-1)}\Big)$ into $\sum_n^\infty c_{nab}, \sum_n^\infty c_{nab+1},...\sum_n^\infty c_{nab+(ab-1)}$ so that it can be managed as the way for the particular example before.

Should I require further that $(ab)$ divides $m$? But even after that,...

Inspired by the proof of Merten's theorem, does $\sum_{i=0}^N\sum_{j=0}^i a_jb_{i-j}=\sum_{j=0}^Na_{N-j}B_j$ helps? Perhaps we can use the value of the series that converges conditionally to prove the Cauchy product above converges? But it is only convergent then, not necessarily absolutely.

However, I thought, even managing the case when $a=b=4$ (the example before) is complicated and heavy, how can I allow $a,b$ to be arbitray (and maybe distinct) natural number and develop a general algorithm to prove the theorem?

If there is something like mathematical induction to handle countable infinite amount of statement, which is applicable here, maybe I can handle it. Well... however, induction obviously doesn't seems to work here directly. (or maybe, there is, an exciting, trick?)

I observed that in the first example stated in the paper, $\sum_{n=0}^\infty c_n=0$, which is because $\big(\sum_{n=0}^\infty a_n\big)\big(\sum_{n=0}^\infty b_n\big)=0$. Should we make more conditions (such as requiring $\sum_{n=0}^\infty u_n=0$?) according to this?

Any hope will be appreciated. I may just want some constructive hints... Thank you!