# Calculus Proof

1. Oct 27, 2005

### kreil

I have this problem on my HW assignment and I just need a little help getting started...

Let {an} be a convergent sequence and let $l=\lim_{n{\rightarrow}\infty}a_n$

Prove that if l < p then:

$$\exists N{\in}N \forall n \ge N : a_n < p$$

Thanks!

Last edited: Oct 28, 2005
2. Oct 27, 2005

### Physics Monkey

You have something backwards I think. If L > P then intuitively you expect all the $$a_n$$ to eventually be close to L and thus greater (not less) than P, right? Can you make progress with this intuitive picture in mind?

3. Oct 27, 2005

### hypermorphism

Counterexample: Let an = 1 - (1/(n+1)). Then $\lim_{n\rightarrow\infty}a_n = 1$. Choose p = 1/10 < 1. Then $\forall N \in \mathbb{N} \exists n\ge N : a_n > p$.

4. Oct 28, 2005

### hypermorphism

Not necessarily. Let an = 1/(n+1). Then $\lim_{n\rightarrow\infty} a_n = 0$. If you choose p=9/10 > 0, all the an's near l are less than p. I think a qualifier is missing somewhere.

5. Oct 28, 2005

### Physics Monkey

No, hypermorphism, you missed the condition in kreil's original post that L > P. In your case L < P from which it follows, as you have indicated, the $$a_n$$ will eventually be less than P and near L.

Last edited: Oct 28, 2005
6. Oct 28, 2005

### hypermorphism

Whoops. :rofl: That's probably the right question, then. In that case, kreil should appeal to the definition of convergence.

7. Oct 28, 2005

### kreil

sorry its l < p

kreil

8. Oct 28, 2005

### kreil

I did some work on it and I'm pretty sure I got it...

Show: an<p

l < p $\implies$ p-l > 0

$$l=\lim_{n{\rightarrow}\infty}a_n < p \iff \forall \epsilon > 0 \exists N \in N \forall n \ge N: |a_n-l| < \epsilon$$

Choose e = p - l > 0

$$\exists N \in N \forall n \ge N: |a_n-l|<\epsilon=p-l$$

and thus

$$\exists N \in N \forall n \ge N: a_n<l+{\epsilon}=l+p-l=p$$

Josh

Last edited: Oct 28, 2005
9. Oct 28, 2005

### Physics Monkey

Yeah, great, that makes more sense. Also, your proof looks good.