# Diverging sequence

1. Oct 11, 2008

### sara_87

1. The problem statement, all variables and given/known data

show that the sequence is divergent:
xn: 1,0,1,0,1,0,.....

3. The attempt at a solution

there exists a number p>0 such that there is no N with the property abs(xn-C)<p
where n>N for all n and C is the limit.

I don know how to find this value of p. My teacher said that p must be greater than 1/2 but i cant understand why.
can someone please explain using very basic explanation.

thank you very much.

2. Oct 11, 2008

### JinM

Suppose, for contradiction, that x_n is convergent. Then lim x_n exists, call it C. That is, for any p > 0, we can find N s.t. |x_n - C| < p for all n > N.

It follows then that |x_n - C| < 1/2 and |x_(n+1) - C| < 1/2, for p = 1/2. Can you continue from here? Hint: if x_n = 0, then x_n+1 = 1. Also, use the triangle inequality to find the contradiction.

Last edited: Oct 11, 2008
3. Oct 11, 2008

### tiny-tim

Hi sara_87!

uh? your teacher must mean p < 1/2

whatever C you choose, either 0 or 1 will be further away from C than p, if p < 1/2.

4. Oct 11, 2008

### JinM

Haha yes, noted in my post as well; picking p =1 would yield nothing substantive in the argument.

Is this for an Analysis class, sara_87?

5. Oct 11, 2008

### sara_87

Yes this is for an analysis class.

sorry, i ment p<1/2.
Thanx JinM and tiny-tim.

6. Oct 11, 2008

### sara_87

Hold on, i dont fully understand yet;
lets just say p=3 then why doesnt this work because abs(xn-1)<3

I think im missing the point???????

7. Oct 11, 2008

### JinM

Relax. The sequence diverges, but picking p = 3 won't help you because the limits 1 and 0 are both inside the interval (-3,3), so arguing divergence is useless here. Pick something smaller until one of the limits leave the interval. You're just trying to constrain an infinite number of terms inside a neighborhood, assuming that a limit exist, but the terms keep leaving the interval so there is your contradiction. Just make this more rigorous as I outlined for you above.

8. Oct 11, 2008

### tiny-tim

Yup … you've started trying to prove that it does converge!

9. Oct 11, 2008

### sara_87

ok so lets take p=2/3; why doesnt this work for:
abs(xn-c)<p ??

10. Oct 11, 2008

### JinM

Sara, I'm not an expert on this, as I'm taking the class right now.

It works for p = 2/3; that is, you would leave out the 1 and constrain the 0 inside the interval, but it is not an ideal candidate for a contradiction argument. From the outline above, what you're trying to do is simply argue that

1 = |1-C+C| <= |1-C| + |C| = |1-C| + |0-C| < p + p = 2p.

Come up with a p where 2p < 1. This will give you p <= 1/2. But do you see why this is a contradiction? Don't forget, you're assuming the result holds for all p > 0, but we have a problem here.

11. Oct 11, 2008

If it converges then every subsequence converges and to the same value. Can you find two subsequences that both converge but to different values? No need for $$\delta$$ and $$\varepsilon$$ then.