# Proof of that a limit -> 0 as n-> infinity

1. Aug 21, 2012

### Jesssa

Hey,

I've been trying to work out the following question,

This is it including what I hope is an ok, or on the way to an ok proof.

http://img269.imageshack.us/img269/6156/proofxq.jpg [Broken]

The book has a hint given that states there exists N in naturals such that |z_n|<ε/2 for all n>N so I kind of tried to use that to find a bound on w_n.

Does this look okay?

Jess

Last edited by a moderator: May 6, 2017
2. Aug 21, 2012

### I like Serena

Hi Jesssa!

That looks close...

But why would $|z_1+z_2+...+z_N| < \frac ε 2$?
I think that will usually not be the case.

Suppose $z_i=\frac{10}{i}$ and $ε={1\over 10}$.
I don't think your condition will hold for any N... at all!

Last edited by a moderator: May 6, 2017
3. Aug 21, 2012

### Ray Vickson

Close, but not complete. The point is: (1) you can find K so that the first K terms contribute < ε/2 to sum z/n; and (2) |z_i| ≤ ε/2 for i = K+1, ..., n. What does (2) imply? Hint: it does not imply anything like what you wrote.

RGV

Last edited by a moderator: May 6, 2017
4. Aug 21, 2012

### SammyS

Staff Emeritus
The book's hint plus your observation that
$\displaystyle w_n=\frac{z_1+z_2+z_3+\dots+z_{K}}{n}+\frac{z_{K+1}+\dots+z_n}{n}$​
are both helpful in this proof. Beyond that it seems to me that you are not very close to completing the proof.

What is it that you know?

You know that:
$\displaystyle z_n\ \to\ 0\ \text{ as }\ n\ \to\ \infty$

What does this tell you about finding some integer, K, such that |zk| < ε/2 for all k > K ?​

What is it that you need to show?

You need to show that:
$\displaystyle w_n\ \to\ 0\ \text{ as }\ n\ \to\ \infty$

This is true, if given any ε>0, there exists some positive integer, N, such that |wn| < ε for all n>N.​

5. Aug 22, 2012

### Jesssa

Thanks guys,

i think i got it down now,

the thing that mucked me up was considering the first set of terms,

i asked my teacher and he helped me out =]

Last edited: Aug 22, 2012