How Do You Prove a Sequence Converges Below a Certain Limit?

[kreil]

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


Let {a_n} 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 &lt; p


Thanks!

 

[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?

 

[hypermorphism]

...
Let {a_n} 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 &lt; p
...

Counterexample: Let a_n = 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 &gt; p.

 

[hypermorphism]

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?

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

 

[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.

 

[hypermorphism]

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

 

[kreil]

sorry its l < p

kreil

 

[kreil]

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

Show: a_n<p

l < p \implies p-l > 0

l=\lim_{n{\rightarrow}\infty}a_n &lt; p <br /> <br /> \iff \forall \epsilon &gt; 0 \exists N \in N \forall n \ge N: |a_n-l| &lt; \epsilon

Choose e = p - l > 0

\exists N \in N \forall n \ge N: |a_n-l|&lt;\epsilon=p-l

and thus

\exists N \in N \forall n \ge N: a_n&lt;l+{\epsilon}=l+p-l=p



Josh

 

[Physics Monkey]

sorry its l < p

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