Question

How do you define two divergent sequences whose termwise product is a convergent sequence?

Answer 1

`a_(2k) = 2^(2k)`, `a_(2k+1) = 2^-(4k+2)`

`b_(2k) = 2^-(4k)`, `b_(2k+1) = 2^(2k+1)`

for `k in NN`

Then `a_i * b_i = 2^-i -> 0` as `i -> oo`

Explanation:

If we define

`a_(2k) = 2^(2k)`, `a_(2k+1) = 2^-(4k+2)` for `k in NN`

then the sequence `a_0, a_1, a_2,...` looks like this:

`2^0, 2^-2, 2^2, 2^-6, 2^4, 2^-10, 2^6, 2^-14,...`

This is clearly divergent, with alternate terms increasing exponentially in value.

If we define

`b_(2k) = 2^-(4k)`, `b_(2k+1) = 2^(2k+1)`

then the sequence `b_0, b_1, b_2,...` looks like this:

`2^0, 2^1, 2^-4, 2^3, 2^-8, 2^5, 2^-10, 2^7,...`

Again, clearly divergent.

For each `k in NN`, we have

`a_(2k) * b_(2k) = 2^(2k)*2^-(4k) = 2^-(2k)`

`a_(2k+1) * b_(2k+1) = 2^-(4k+2)*2^(2k+1) = 2^-(2k+1)`

So `a_i*b_i = 2^-i` for all `i in NN`

looks like this:

`2^0, 2^-1, 2^-2,... -> 0` as `i -> oo`