If $\sum_{n\ge 0} a_n$ and $\sum_{n\ge 0} b_n$ are two series, their Cauchy product is defined as $\sum_{n\ge 0} c_n$, where $c_n = \sum^n_{k=0} a_k b_{n-k}$.

As this question points out, finding two conditionally convergent series whose Cauchy product is absolutely convergent is quite hard, but examples do indeed exist. I also learned that the Cauchy product of a divergent series with itself can be absolutely convergent (see here). So do there exists a conditionally convergent series $\sum_{n\ge 0} a_n$ such that the Cauchy product of $\sum_{n\ge 0} a_n$ with itself is absolutely convergent?

  • $\begingroup$ Interesting question! I thought I might be able to use the binomial series, but the binomial series can't serve as an example due to conflicting requirements ($2\alpha\in\mathbb{Z}_{\ge 0}$ and $-1 < \text{Re}(\alpha) \le 0$). $\endgroup$ May 6, 2022 at 7:02
  • $\begingroup$ If $f(z)=\sum_{n=0}^{\infty}a_nz^n$, then $f(z)$ must have radius of convergence equal to exactly $1$. This implies that $\limsup_{n \to \infty} \sqrt[n]{|a_n|}=1$ and that the series for $f(z)^2$ has radius of convergence at least equal to $1$ (which is necessary for absolute convergence of $\sum_{n \ge 0} c_n$). $\endgroup$ May 7, 2022 at 8:05
  • $\begingroup$ @VarunVejalla Yes, and I was wondering if more can be said about the properties of $f(z)$ :) $\endgroup$ May 8, 2022 at 10:11
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    $\begingroup$ Looking at the taylor series of $\sqrt{x+1}$ at $0$, I would expect divergence for $x=-1$ and convegence for $x=1$, so at $x=1$ its conditionally convergent. But $\sqrt{x+1}^2=x+1$ should have globally convergent taylor series, so enough stuff will cancel out to get absolute convergence if you square the taylor series of $\sqrt{x+1}$. Atleast thats my expectation, maybe the details don't work out. $\endgroup$
    – s.harp
    Jun 25, 2022 at 11:23
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    $\begingroup$ @s.harp Sorry but the taylor series of $\sqrt{x+1}$ converge absolutely at $x=-1$. In fact, I would not be astonished if the example I'm looking for does not exist at all. $\endgroup$ Jun 25, 2022 at 12:34


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