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?

  • $\begingroup$ 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 ... $\endgroup$ – Hagen von Eitzen Jul 6 '15 at 7:54
  • $\begingroup$ Say I had numerical series $\sum a_n$ and $\sum b_n$. If I wanted to compute the product, could I say $(\sum a_n)(\sum b_n) = a_0 \sum b_n + a_1 \sum b_n + \ldots$? $\endgroup$ – user217285 Jul 6 '15 at 7:56
  • $\begingroup$ Yes you are treating $\sum b_n$ as a constant here. In a numerical application you would use unnecessarily many multiplications though ... $\endgroup$ – Hagen von Eitzen Jul 6 '15 at 7:57

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.

  • $\begingroup$ Sir, how do you get inequality $(k+1)(n-k+1) \ge (n/2+1)^2$? can you elaborate please. $\endgroup$ – Akash Patalwanshi Dec 3 '19 at 12:38
  • 1
    $\begingroup$ You can study this as a function of $k$ and find the minimum. $\endgroup$ – mathcounterexamples.net Dec 3 '19 at 12:42
  • $\begingroup$ Sir can you elaborate more... $\endgroup$ – Akash Patalwanshi Dec 3 '19 at 12:46

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