[Math] Can the Cauchy product of divergent series with itself be convergent

divergent-seriesreal-analysissequences-and-series

For series $\sum a_n$ and $\sum b_n$ their Cauchy product is the series $\sum c_n$ where
$$
c_n = a_0b_n+a_1b_{n-1}+\ldots+a_nb_0.
$$
Does there exist a sequence $\sum a_n$ such that:

  1. $\sum a_n$ is divergent?
  2. The Cauchy product of $ \sum a_n$ and $\sum a_n$ is convergent?

Best Answer

Let $c_0=1$, $c_1=-2$, $c_n=0$ for $n>1$. Let $a_0=1$, and for $n\ge1$ recursively $$a_n=\frac12\left(c_n-\sum_{k=1}^ {n-1}a_ka_{n-k} \right).$$ Then clearly $\sum c_n=\sum a_n\cdot \sum a_n$ in the sense of Cauchy product, and $\sum c_n$ is of course very convergent. Assume $\sum a_n$ converges. Then $\sum a_n x^n$ converges absolutely on $(-1,1)$. By absolute convergence, $\sum c_n x^n=(\sum a_n x^n)^2$ for $|x|<1$, which is absurd as $\sum c_n x^n<0$ for $x>\frac12$.

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