# Sequence limits?

1. Sep 9, 2009

### dannysaf

Let lim n →∞ XnYn = 0. Is it true that Limn →∞ Xn= 0
or Limn →∞Yn = 0 (or both)?

2. Sep 9, 2009

### njama

lim n →∞ XnYn = lim n →∞ Xn * lim n →∞ Yn

One or even both limits need to be 0 so that lim n →∞ Xn * Yn =0

I guess you are right.

3. Sep 9, 2009

### VietDao29

Well, this is only true when both xn, and yn have limits. So, what if they don't? Say, what if they're oscillating?

4. Sep 9, 2009

### slider142

This equation is only correct if the individual limits on the right-hand side exist. The proposition in the original post is false and can be disproven with a counterexample.

5. Sep 10, 2009

### njama

Yup, sorry I forgot to mention that the series must converge, in other way they would be divergent and we could not separate them.

6. Sep 10, 2009

### fmam3

Simple counterexample. Clearly, $$\lim 1/n = \lim $$n \cdot 1/n^2$$ = 0$$ but we have that $$\lim n = +\infty$$ and $$\lim 1/n^2 = 0$$.

7. Sep 10, 2009

### Elucidus

This example satisfies the original claim that at least one of the sequences converges to 0. A counter example for this must consist of two sequences whose product vanishes but both sequence do not converge to 0.

--Elucidus

8. Sep 11, 2009

### VietDao29

You've misquoted yet again, Elucidus..

You shouldn't forget to wear your glasses, then..

9. Sep 11, 2009

### Elucidus

Indeed. I'm surprised. I was trying to quote this post:

To which my original comment makes more sense. Sorry for any confusion.

--Elucidus

10. Sep 11, 2009

### g_edgar

Let $x_n$ be alternately 1 and 0, let $y_n$ be alternately 0 and 1.
Then $x_n y_n = 0$ for all n, but neither individual limit exists. In particular, neither individual limit is zero.

11. Sep 11, 2009

### HallsofIvy

Staff Emeritus
Excellent example. However, neither $x_n$ nor $y_n$ converges and the O.P. finally told us.

12. Sep 11, 2009

### njama

HallsofIvy I am not the O.P, mate. I was just adding, additional explanation of my first post.