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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?

Jianing Song
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  • 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 $-1 < \text{Re}(\alpha) \le 0$). – Varun Vejalla May 06 '22 at 07:02
  • 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 $\sum_{n \ge 0} c_n$). – Varun Vejalla May 07 '22 at 08:05
  • @VarunVejalla Yes, and I was wondering if more can be said about the properties of $f(z)$ :) – Jianing Song May 08 '22 at 10:11
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    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. – s.harp Jun 25 '22 at 11:23
  • Annoyingly $\sum_{n=0}^\infty (-1)^{n-1} \binom{1/2}{n} = -1 + 1/2 + 1/8 + 1/16 + 5/128 + \cdots $ converges to $0$. (Proof: the upper bound on $\binom{2n}{n}$ in https://en.wikipedia.org/wiki/Central_binomial_coefficient implies $|\binom{1/2}{n}| = \frac{1}{2n} |\binom{-1/2}{n-1}| = \frac{1}{2n4^{n-1}} \binom{2(n-1)}{n-1} \le \frac{1}{2n 4^{n-1}} \times \frac{4^n}{\sqrt{\pi n}} = \frac{2}{n\sqrt{n}}$, and now use the comparison test to get convergence, then use Abel's Lemma.) So this is an example of the Cauchy square of an absolutely convergent series being absolutely convergent. – Mark Wildon Jun 25 '22 at 12:01
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    @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. – Jianing Song Jun 25 '22 at 12:34
  • Have you tried looking for a function that is integrable but not absolutely integrable, but is absolutely integrable after convolution with itself? Maybe there's a result about that that carries over to the cauchy product. – dirich1337 Jun 29 '22 at 08:41

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