"Convoluted" sum convergence: $\sum_{i=0}^{n} a_i \sum_{k=0}^{n-i} b_k$

Question

Suppose $\sum_{j=0}^{\infty} a_j =a$, $\sum_{j=0}^{\infty} b_j =b$. Is it true (or under what conditions is it true) that:

$$\lim_{n \rightarrow \infty}\sum_{i=0}^{n} \left( a_i \sum_{k=0}^{n-i} b_k \right) = ab \ ? \tag{1}$$

This arose in one of my earlier questions (here) and at first I thought it was sort of obvious - expanding it out, but now I'm confused again and I want to make this rigorous. I've looked into Cauchy's product formula since it rang a bell but I can't really make my sum look like anything similar to it. One (perhaps trivial) thing I noticed about this is the symmetry:

$$\sum_{i=0}^{n} \left( a_i \sum_{k=0}^{n-i} b_k \right) = \sum_{i=0}^{n} \left( a_k \sum_{k=0}^{n-i} b_i \right)$$

But doesn't get me anywhere. What I'm looking for is a rigorous proof of $(1)$, and I can't make any headway on that. Any help would be greatly appreciated.

Answer 1

Since $$\sum_{i=0}^{n} \left( a_i \sum_{k=0}^{n-i} b_k \right) = \sum_{i=0}^{n} \left( a_k \sum_{k=0}^{n-i} b_i \right)=\sum _{i+k\le n}a_ib_k$$

What you are considering is exactly the Cauchy's product formula.

Answer 2

This is not true in general. As Ma Ming has pointed out, your sum is a partial sum of the Cauchy product of the two sequences. These partial sums can be shown to converge to $ab$ if at least one of the series converges absolutely. Generally the partial sums of the Cauchy product need not converge; a counterexample is afforded by $a_n=b_n=(-1)^n/\sqrt{n+1}$, both of which yield conditionally convergent series, whereas the partial sums of the Cauchy product do not converge.

See also Wikipedia.