# Sequence convergence?

1. Sep 24, 2009

### tarheelborn

1. The problem statement, all variables and given/known data
If {s_n} is a bounded sequence of real numbers and {t_n} converges to 0, prove that {s_n*t_n} converges to 0.

2. Relevant equations

3. The attempt at a solution
I know that since s_n is bounded, ||s_n|| <= M, but I cannot seem to find a way to make this work into the proof for convergence. Any help will be appreciated.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 24, 2009

### LCKurtz

Since you have |sn| ≤ M what can you say about |sntn|?

3. Sep 24, 2009

### tarheelborn

You can conclude that ||s_n*t_n||<=||t_n|*M. I know that t_n < epsilon, but I am having trouble working around to that.

4. Sep 24, 2009

### LCKurtz

OK. Can you use that and the fact you know |tn| → 0, to claim if you were given ε > 0, you can make

|tnsn| < ε ?

5. Sep 24, 2009

### tarheelborn

Thanks; you have helped immensely.

6. Sep 24, 2009

### tarheelborn

Sorry, but I must still be stumped. I wrote the following:
Proof: Let \epsilon > 0. By the definition of bounded, |s_n|<=M, where M is a real number and n is an integer. This implies that |s_n*t_n|<=|t_n|*M. Since we know that t_n approaches 0 and \epsilon > 0, it follows that |s_n*t_n| < \epsilon. We can therefore conclude that s_n*t_n approaches 0. End of proof.

I can't seem to get the "we need to find a positive integer N such that |s_n*t_n|<\epsilon" part. I can't find N. I seem to be caught up in the abstractness of it. I will certainly appreciate just a tiny bit more help. Thank you.

7. Sep 24, 2009

### fmam3

I think you have most of it. First of all, if $$\lim t_n = 0$$, then what can you say about it? So, let $$\forall \varepsilon > 0$$. Since $$(s_n)$$ is bounded, then $$\exists M > 0$$ such that $$|s_n| \leq M$$ for $$\forall n \in \mathbb{N}$$. Since $$\lim t_n = 0$$, $$\exists N$$ such that $$\forall n > N$$, we have $$|t_n - 0| < \varepsilon /M$$. Note that we have $$\varepsilon / M > 0$$ and this is perfectly valid (and the key to why your proof is a little lacking --- can you think of why we can write $$\varepsilon / M$$ instead of just the usual $$\varepsilon$$ on the right hand side of the above?).

Then it follows that
$$|s_n t_n - 0| \leq M|t_n| < M \cdot \frac{\varepsilon}{M} = \varepsilon$$, for $$\forall n > N$$. This completes the proof. :)

8. Sep 24, 2009

### tarheelborn

Ah, yes, it does!!! Thank you so much. This time I really do have it! Have a nice evening and thank you.