Diverging Sequence: Proving p>0 for 1, 0, 1, 0, 1, 0...

[sara_87]

Homework Statement

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

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 can't understand why.
can someone please explain using very basic explanation.

thank you very much.

 

[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.

 

[tiny-tim]

show that the sequence is divergent:
xn: 1,0,1,0,1,0,...
…
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 can't understand why.

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.

 

[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?

 

[sara_87]

Yes this is for an analysis class.

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

 

[sara_87]

Hold on, i don't fully understand yet;
lets just say p=3 then why doesn't this work because abs(xn-1)<3

I think I am missing the point??

 

[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.

 

[tiny-tim]

I think I am missing the point??

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

 

[sara_87]

ok so let's take p=2/3; why doesn't this work for:
abs(xn-c)<p ??

 

[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.

 

[statdad]

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.

 

[sara_87]

Oh ok, i see the point now. we can't find a value p which is > 1/2 that would satisfy our property abs(xn-C) so that means there is no limit.

:)
Thanks all.