# In search of a counter-example

Quantumpencil

## Homework Statement

Is it true that if {pn}{qn}-> 0, that either pn or qn converges to 0?

## The Attempt at a Solution

So far I've made an argument for why it doesn't have to follow (Because we don't know that both sequences converge, and we could have two divergent sequences or a divergent sequence and a convergent sequence whose product converges, for example the alternating harmonic series), However I can't think of any particular counter-example where the product converges to 0.

Citan Uzuki
Try to arrange it so that the even terms of {p_n} converge to zero, but {p_n} does not, and conversely, the odd terms of {q_n} converge to zero, but {q_n} does not.

Quantumpencil
That can't happen though can it? If {p_n}(even)-> 0, then the even terms form a subsequence, so wouldn't the sequence have to converge to 0 also?

Quantumpencil
nevermind, I got it I think.

(1,0,1,0,1...)
(0,1,0,1,0...)

My concern isn't relevant because they diverge.

Homework Helper
Yes, if {x_n} converges to x and {y_n} converges to y, then {x_ny_n} must converge to xy. So if {x_n} and {y_n} converge AND {x_ny_n} converges to 0, either x or y must be 0.

But you can find sequences that do NOT converge but {x_ny_n} converges to 0.

Quantumpencil
Hm... what about if {x_ny_n} converges, but only {x_n} or {y_n} converges?

Would that be possible, or would {x_n}-> x imply {y_n} also converged as well?

Homework Helper
n*(1/n) converges, right?

Citan Uzuki
Hm... what about if {x_ny_n} converges, but only {x_n} or {y_n} converges?

Would that be possible, or would {x_n}-> x imply {y_n} also converged as well?

As dick showed, it is possible to have {x_n*y_n} and {x_n} converge, but {y_n} diverge. However, this is only possible if $x_n \rightarrow 0$. Because if $x_n y_n \rightarrow L$ and $x_n \rightarrow x \neq 0$, then applying the main limit theorem shows that:

$$\lim y_n = \lim \frac{x_n y_n}{x_n} = \frac{\lim x_n y_n}{\lim x_n} = \frac{L}{x}$$

i.e. $y_n \rightarrow \frac{L}{x}$