Can You Prove lim(a_n b_n) = 0 for a Bounded Sequence b_n and lim(a_n) = 0?

[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\inN 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

 

[Petek]

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

Observe that |a_nb_n| = |a_n| |b_n|. Also, |b_n| &lt; 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?

 

[Punkyc7]

do you just say for any b_{n} choose an a_{n} such that b_{n} * a_{n} <\epsilon

 

[vela]

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\inN and we also know that |a_{n}|< \epsilon for all \epsilon>0.

The bolded part isn't correct. What \lim_{n \to \infty} a_n = 0means is that given \varepsilon_1 &gt; 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 a_nb_n sequence to \varepsilon_1 to get what you need for the proof.

 

[Petek]

You must show that, given any \epsilon &gt; 0 that there exists an N > 0 such that if n> N, then |a_nb_n| &lt; \epsilon. Now you know that (a_n) converges to 0. That implies that, given any \epsilon &gt; 0, then there exists an N > 0 such that |a_n| &lt; \epsilon if n > N. Does that N work? No, because if n > N then |a_nb_n| = |a_n| |b_n| &lt; \epsilon M, but we needed |a_nb_n| &lt; \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.

 

[Punkyc7]

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

Could you just define \epsilonM 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| &lt; (\epsilon/M) *M=\epsilon,

 

[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| &lt; \epsilon. Now, since (a_n) converges to 0, there exists an N_1 &gt; 0 such that if n &gt; N_1, then |a_n| &lt; \epsilon/M. Therefore, if n &gt; N_1, then |a_nb_n| &lt; (\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| &lt; \epsilon. Hope that makes sense.