# Homework Help: Sequence proof

1. Sep 2, 2011

### Punkyc7

If (b$_{n}$) is a bounded sequence ad lim(a$_{n}$)=0 show that lim(a$_{n}$b$_{b}$) =0

Pf/

Let b$_{n}$ be bounded and the lim(a$_{n}$)=0. Since b$_{n}$ is bounded we know that $\exists$ a real number M $\ni$ |b$_{n}$|<M for all n$\in$$N$ and we also know that |a$_{n}$|< $\epsilon$ for all $\epsilon$>0.

My problem is how do I go from here. I don't believe you can say that the lim(b)*lim(a)=lim(ab)=0 because we don't know what the lim(b) is

2. Sep 2, 2011

### Petek

Here's how to start: Let $\epsilon > 0$. Then we have to show that there exists an N > 0 such that if n > N, then $|a_nb_n| < \epsilon$.

Observe that $|a_nb_n| = |a_n| |b_n|$. Also, $|b_n| < M$ for all n and we can "make $|a_n|$ as small as we like" for sufficiently large n. Can you put the pieces together from here?

3. Sep 2, 2011

### Punkyc7

do you just say for any b$_{n}$ choose an a$_{n}$ such that b$_{n}$ * a$_{n}$ <$\epsilon$

4. Sep 2, 2011

### vela

Staff Emeritus
The bolded part isn't correct. What $$\lim_{n \to \infty} a_n = 0$$means is that given $\varepsilon_1 > 0$, there exists $N \in \mathbb{N}$ such that n>N implies $|a_n|\lt \varepsilon_1$.

Think about how you might relate $\epsilon$ for the anbn sequence to $\varepsilon_1$ to get what you need for the proof.

5. Sep 2, 2011

### Petek

You must show that, given any $\epsilon > 0$ that there exists an N > 0 such that if n> N, then $|a_nb_n| < \epsilon$. Now you know that $(a_n)$ converges to 0. That implies that, given any $\epsilon > 0$, then there exists an N > 0 such that $|a_n| < \epsilon$ if n > N. Does that N work? No, because if n > N then $|a_nb_n| = |a_n| |b_n| < \epsilon M$, but we needed $|a_nb_n| < \epsilon$. Can you find a way to modify this argument to get the desired result? Perhaps your text or lecture notes contain examples that might help.

6. Sep 2, 2011

### Punkyc7

That implies that, given any $\epsilon > 0$, then there exists an N > 0 such that $|a_n| < \epsilon$ if n > N. Does that N work? No, because if n > N then $|a_nb_n| = |a_n| |b_n| < \epsilon M$, but we needed $|a_nb_n| < \epsilon$.

Could you just define $\epsilon$M to be the new $\epsilon$?

Or can you say let $\epsilon$>0, Since M is a real number>0 we know $\epsilon$/M>0 so $|a_nb_n| = |a_n| |b_n| < (\epsilon/M) *M=\epsilon$,

Last edited: Sep 2, 2011
7. Sep 3, 2011

### Petek

The last sentence in your post is basically correct. You just have to phrase it differently. Since you have the right idea, here's the way I would phrase it:

Let $\epsilon$ be any number > 0. We have to show that there exists an N > 0 such that if n > N, then $|a_nb_n| < \epsilon$. Now, since $(a_n)$ converges to 0, there exists an $N_1 > 0$ such that if $n > N_1$, then $|a_n| < \epsilon/M$. Therefore, if $n > N_1$, then $|a_nb_n| < (\epsilon/M)(M) = \epsilon$, as required.

The point is that we define a different "$\epsilon$" for $a_n$ so that we get $|a_nb_n| < \epsilon$. Hope that makes sense.