Does the Sequence Sn = {0 if n is even, -1 if n is odd} Converge?

[missavvy]

Homework Statement

Suppose S_n is a sequence defined by:

S_n = 0, if n is even
-1, if n is odd.

Show Sn does not converge

Homework Equations

The Attempt at a Solution

By contradiction:

Suppose there exists an x>0 s/t for all natural numbers N, there exists n>=N which implies |s-s_n| >= x

I am trying to find an x but for some reason it doesn't ...
For example if I take x = 1, and let s = the limit of s_n

N+1 and N+2 are > than N so

|s_N+1 - s| < 1
|s_N+2 - s| < 1

|s_N+1 - s_N+2| = 1
1 = |s_N+1 - s + s - s_N+2|
<= |s_N+1 - s| + |s - s_N+2|
< 1+1
< 2
which is true and not a contradiction! what am I doing wrong? :S

 

[╔(σ_σ)╝]

Show that the sequence has two subsequential limit.

 

[missavvy]

ok... stupid question, but, why does this mean it diverges?
Because can a sequence not have more than 1 subsequence which converges? >_<
like i know it diverges but i just don't understand the proof...

 

[╔(σ_σ)╝]

A sequence can have more than one subsequence that converges but if the original sequence converges the subsequential limits have to be the same.

If a sequence have two subsequencial limits that are different it cannot converge. Consider the sequence
-1,1,-1,1...
You can find two subsequences with different limits and it is obvious that the sequence does not converge

It would help to think about the definition of a sequence converging. In other for a sequence to converge all points or the sequence aftet some n_0 must be very very close to each other.

In your case there are only two accumulation points which are the only candidates for the limit. Assuming that either 1 or 0 is the limit and picking epsilon to be 1/2 leads to contradiction since far out in the sequence you have a 1 or 0 somewhere.

 

[missavvy]

Ohhh that makes a lot of sense (thanks for explaining it in english!), thank you so much! :D

 

[HallsofIvy]

In particular, "diverge" does not mean "goes to infinity". It simply means "does not converge". You can give an indirect proof that this series does not converge using the definition of "limit". Suppose the sequence conveges to some limit, "L". Then given any \epsilon&gt; 0, there must exist N such that if n> N, then |a_n- L|&lt; \epsilon.

Since a_n is always 0 or 1, take \epsilon to be any positive number less than 1/2. Then if L< 1/2 itself, |1- L|> 1/2. If L> 1/2, then |0- L|> 1/2. If L= 1/2, both |1- L|= |1- 1/2| and |0- L|= |0- 1/2| are equal to 1/2 and so never less than \epsilon.