# Proving a limit and the Order Limit Theorem

1. Feb 13, 2008

### rbzima

If $$(a_n) \rightarrow 0$$ and $$\left|b_n - b\right| \leq a_n$$, then $$(b_n) \rightarrow b$$.

I'm a little bit stuck on proving this! Part (iii) of the Order Limit Theorem states:

Assume:$$lim a_n = a$$ and $$lim b_n = b$$
If there exists $$c \in$$ R for which $$c \leq b_n$$ for all $$n \in$$ N, then $$c \leq b$$. Similarly if $$a_n \leq c$$ for all $$n \in$$ N, then $$a \leq c$$.

I can then take this to a point where $$b_n = b$$, but somehow that doesn't seem right because $$\left|b_n - b\right|<\epsilon$$, where $$\epsilon > 0$$. Where am I going wrong here!

2. Feb 13, 2008

### d_leet

What happens if you choose n so large that |an| is less than epsilon? Why can you choose such an n?

3. Feb 13, 2008

### rbzima

That's exactly what I want. I want |an| to be less than epsilon because it implies convergence for any n, which means |an| is practically nothing because it approaches 0.

I'm not really sure what you are getting at here.

4. Feb 13, 2008

### d_leet

You know {an} converges to 0, what does that tell you?

5. Feb 13, 2008

### rbzima

$$b_n - b = 0$$

So, basically, $$b_n = b$$

But I'm wondering if that's really all thats necessary?

6. Feb 13, 2008

### d_leet

I don't understand how you get that from what I wrote.

Now IGNORE {bn} for a minute you know that {an} converges to zero, what does that mean?

7. Feb 13, 2008

### rbzima

We had $$a_n$$ in the inequality, that's what I thought you were referring to.

Basically, the sequence approaches zero! Convergent series are bounded above and below, and a sequence converges when, for every positive integer $$\epsilon$$, there exists and $$N \in$$ N such that whenever $$n \leq$$ N, it follows that |an - a| < $$\epsilon$$.

I know the definition of convergence, in this case: |an| < $$\epsilon$$

Shall I list the properties of convergent sequences while I'm at it too! LOL!

8. Feb 13, 2008

### sutupidmath

If you are trying to prove just this part, than here you go:
since $$(a_n) \rightarrow 0$$ it means that for every e>o there exists some N(e)>0 such that whenever n>N we have Ia_nI<e, which actually is

-e<a_n<e
Now going back to $$\left|b_n - b\right| \leq a_n<e$$ which actually means that b_n converges to b.

what else are u looking for?
P.S. This is what d_leet also suggested, i guess!

9. Feb 13, 2008

### d_leet

Two things. One please don't post entire solutions to questions such as this. And two, there actually needs to be an additional restriction to {an} or to the inequality of |b{sub]n[/sub]-b|<an in order for it to make sense; that restriction being either {an} is a sequence of positive terms or the inequality should be |bn-b|<|an|