Non-Cauchy products for series?

Question

Let $\sum a_n$ and $\sum b_n$ be two absolutely convergent series. My question is

Is it possible to take the product of the two series $(\sum a_n)(\sum b_n)$ and get a result different from the Cauchy product? In other words, is the product of two series well-defined? What if $\sum a_n$ and $\sum b_n$ are conditionally convergent?

My motivation for the question is as follows. Cauchy says for two series $\sum a_n$ and $\sum b_n$, we should write

\begin{align*} \left(\sum a_nx^n\right)\left(\sum b_nx^n\right) &= (a_0 + a_1 x + a_2 x^2 + \cdots)(b_0 + b_1 x + b_2 x^2 + \cdots) \\ &= a_0b_0 + (a_0b_1 + a_1b_0)x + (a_0b_2 + a_1b_1 + a_2b_0)x^2 + \cdots \\ &= \sum_{n=0}^\infty \left(\sum_{k=0}^n a_kb_{n-k}\right) x^n \end{align*} and set $x = 1$ to get the desired formula.

But this does some sneaky re-arrangements (I believe). If both $\sum a_n$ and $\sum b_n$ are absolutely convergent, I think we should have no problems in re-arrangements, but I am not sure. So why can't we write

\begin{align*} \left(\sum a_nx^n\right)\left(\sum b_nx^n\right) &= (a_0 + a_1 x + a_2 x^2 + \cdots)(b_0 + b_1 x + b_2 x^2 + \cdots) \\ &= a_0(b_0 + b_1 x + b_2 x^2 + \cdots) + a_1x(b_0 + b_1 x + b_2 x^2 + \cdots) + \cdots \\ &= a_0 \left(\sum b_n\right) + a_1x \left(\sum b_n\right) + \cdots \\ &= \sum_{i=0}^\infty a_ix^i \left(\sum b_nx^n\right) . \end{align*}

Would those lead to the same result?

Answer 1

I answer here to your question highlighted in yellow.

According to Mertens theorem, if one of the two series converges absolutely then the Cauchy product of the two series converges to the product of the sum of the two series.

If both series are conditionally convergent this might not be the case.

For example take $$a_n=b_n=\frac{(-1)^n}{\sqrt{n+1}}$$ Denoting $\sum c_n$ the Cauchy product, you have $$c_n=\sum_{k=0}^n \frac{(-1)^k}{\sqrt{k+1}} \frac{(-1)^{n-k}}{\sqrt{n-k+1}} = (-1)^n\sum_{k=0}^n \frac{1}{\sqrt{(k+1)(n-k+1)}}$$ and $(k+1)(n-k+1) \ge (n/2+1)^2$ so $$\vert c_n \vert \ge \frac{n+1}{n/2+1}$$ and the RHS of the inequality converges to $2$. Therefore the Cauchy product does not converge.