Two divergent series such that their product is convergent I faced a series question it goes something like give an example of 2 divergent series such that when the 2 series are multiplied to each other, the new series becomes convergent, although it looks absurdly simple still am at a loss.
 A: A pair of divergent sequences $$a_n = \begin{cases}0 & \text{for } n \text{ odd} \\ 1 & \text{for } n \text{ even}\end{cases} \quad \text{ and } \quad b_n = 1-a_n$$
makes as a 'product' a convergent sequence $$a_n\cdot b_n = 0$$
and similary their series: while $\sum\limits_n{a_n} = \sum\limits_n{b_n} = \infty$ the 'product' is $$\sum\limits_n{a_n\cdot b_n} = 0$$
A: It depends on what you mean by multiply.. If you're looking for series $a_n$, $b_n$ such that 
$\sum a_n$ diverges, $\sum b_n$ diverges, but $\sum a_nb_n$ converges, then you can just take $a_n = \frac 1n = b_n$.
Another way to intend multiplication is to use Cauchy's product, but that requires that the two series to converge and at least one of them do so absolutely, so I guess this is not what you're talking about :-)
Edit: 
What I said about the cauchy product is not correct: If those conditions are met, then the product converges, but it's not a necessary condition. Refer to Daniel Fisher comment below ;-)
A: $a_n = (-1)^n = b_n$. Both are divergent but product is {1,1,1....} clearly converges to 1. 
