Proving Convergence of Bounded Sequence {s_n*t_n} with Limit 0

[tarheelborn]

Homework Statement

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.

Homework Equations

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.

 

[LCKurtz]

Since you have |s_n| ≤ M what can you say about |s_nt_n|?

 

[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.

 

[LCKurtz]

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

|t_ns_n| < ε ?

 

[tarheelborn]

Thanks; you have helped immensely.

 

[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.

 

[fmam3]

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.

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 &gt; 0. Since (s_n) is bounded, then \exists M &gt; 0 such that |s_n| \leq M for \forall n \in \mathbb{N}. Since \lim t_n = 0, \exists N such that \forall n &gt; N, we have |t_n - 0| &lt; \varepsilon /M. Note that we have \varepsilon / M &gt; 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| &lt; M \cdot \frac{\varepsilon}{M} = \varepsilon, for \forall n &gt; N. This completes the proof. :)

 

[tarheelborn]

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