# Non-Cauchy products for series?

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?

• Yes, you can always transform $(\sum a_n) B=\sum (a_n B)$. But here that would give you a series of power series, not a power series ... – Hagen von Eitzen Jul 6 '15 at 7:54
• Say I had numerical series $\sum a_n$ and $\sum b_n$. If I wanted to compute the product, could I say $(\sum a_n)(\sum b_n) = a_0 \sum b_n + a_1 \sum b_n + \ldots$? – user217285 Jul 6 '15 at 7:56
• Yes you are treating $\sum b_n$ as a constant here. In a numerical application you would use unnecessarily many multiplications though ... – Hagen von Eitzen Jul 6 '15 at 7:57

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.
• Sir, how do you get inequality $(k+1)(n-k+1) \ge (n/2+1)^2$? can you elaborate please. – Akash Patalwanshi Dec 3 at 12:38
• You can study this as a function of $k$ and find the minimum. – mathcounterexamples.net Dec 3 at 12:42