Proof of Non-Cauchy Sequence: s_n = 1 + 1/2 + ... + 1/n

In summary: So here's what I've got:In summary, we are trying to prove that the sequence {s_n} is not Cauchy. We can choose any epsilon we want, but then our opponent gets to choose N. If we can always find m and n greater than N such that | s_m - s_n | > epsilon, then we win and the sequence is not Cauchy. However, if our opponent can always choose an N that beats us, then the sequence is Cauchy.
  • #1
tarheelborn
123
0

Homework Statement



For each n \in N, let s_n = 1 + 1/2 + ... + 1/n. By considering s_2n - s_n, prove that {s_n} is not Cauchy.

Homework Equations





The Attempt at a Solution



I know that s_2n - s_n = (1 + 1/2 + ... + 1/n + 1/(n+1) + ... + 1/2n) - (1 + 1/2 + ... + 1/n)
= (1/(n+1) + 1/(n+2) + ... + 1/2n
> 1/2n + 1/2n + ... + 1/2n
= n * 1/2n = 1/2

Let \epsilon > 0. Then I need to find N \in N such that m, n >= N. So if I let N = 1/2...

And that's where I lose it. These don't work like regular epsilon proofs!
 
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  • #2
Wait, are you sure of what you need to find? For any epsilon you need to find an N such that IF m,n > N, THEN | s_m - s_n | < epsilon.

You just proved that for any m, s_2m is more than 1/2 away from s_m. So you win. Just choose epsilon < 1/2.
 
  • #3
Thank you so much; I think that was definitely part of my problem. I am also a bit unsure how to write it up in a formal way. Given what I already had, would the following finish it off?

Let 0 < epsilon < 1/2. Then, for any m, n >=N, |s_2n - s_n| >= 1/2. Therefore {s_n} is not Cauchy.

And I don't really need to say anything about the value of N, right?
 
  • #4
Well you should say something like "for any N" since you haven't identified N in your proof. Basically you are saying: no matter what N I choose, this relationship always exists for m,n greater than N. Thus NO N satisfies the condition for the given epsilon, hence the sequence isn't Cauchy.

In doing these proofs I often like to think of it in an adversarial way. I'm trying to prove that a sequence isn't Cauchy. So now I get to pick an epsilon. Then my opponent, a very mean man who hates undergraduate math students, gets to pick a number N. Then I get to pick m and n > N. If | s_m - s_n | > epsilon, I win. If not, I lose.

Now if I can play this game such that my opponent can never win, then the sequence isn't Cauchy. On the contrary, if he can always respond to my pick of some epsilon with an N that beats me, then the sequence is Cauchy.

I'm a game theory kind of guy though, so maybe others prefer to just "for any" and "there exists".
 
  • #5
You should show:

[tex]\exists\ \epsilon = (put\ here\ the\ right\ value)\ so\ \forall\ N\ there\ are\ m>n>N\ so\ that:\ |\sum_{k=n+1}^{m}a_k|\geq\epsilon [/tex]

The most important thing is to understand and "feel" definitions and their meaning. You can't progress further before fully grasping them.
hgfalling said:
...
In doing these proofs I often like to think of it in an adversarial way. I'm trying to prove that a sequence isn't Cauchy. So now I get to pick an epsilon. Then my opponent, a very mean man who hates undergraduate math students, gets to pick a number N. Then I get to pick m and n > N. If | s_m - s_n | > epsilon, I win. If not, I lose.
...

This one is good :rofl:
 
Last edited:
  • #6
hgfalling said:
Well you should say something like "for any N" since you haven't identified N in your proof. Basically you are saying: no matter what N I choose, this relationship always exists for m,n greater than N. Thus NO N satisfies the condition for the given epsilon, hence the sequence isn't Cauchy.

In doing these proofs I often like to think of it in an adversarial way. I'm trying to prove that a sequence isn't Cauchy. So now I get to pick an epsilon. Then my opponent, a very mean man who hates undergraduate math students, gets to pick a number N. Then I get to pick m and n > N. If | s_m - s_n | > epsilon, I win. If not, I lose.

Now if I can play this game such that my opponent can never win, then the sequence isn't Cauchy. On the contrary, if he can always respond to my pick of some epsilon with an N that beats me, then the sequence is Cauchy.

I'm a game theory kind of guy though, so maybe others prefer to just "for any" and "there exists".

Thanks so much for your analogy! That really helps a lot.
 

What is a Cauchy sequence?

A Cauchy sequence is a sequence of numbers in which the terms get closer and closer together as the sequence progresses. In other words, for any small number, there exists a point in the sequence where all subsequent terms are within that distance from each other.

What is the importance of proving a sequence is Cauchy?

Proving that a sequence is Cauchy is important because it guarantees that the sequence has a limit. This is because the definition of a Cauchy sequence is closely related to the definition of a convergent sequence, where the limit is the value that the terms of the sequence approach.

How is a Cauchy sequence proof typically done?

A Cauchy sequence proof is typically done using the Cauchy criterion, which states that a sequence is Cauchy if and only if, for any small number, there exists a point in the sequence where all subsequent terms are within that distance from each other. This proof often involves using the properties of real numbers and the definition of a limit.

What is the difference between a Cauchy sequence and a convergent sequence?

A Cauchy sequence is a sequence of numbers in which the terms get closer and closer together, whereas a convergent sequence is a sequence of numbers that has a limit. All convergent sequences are Cauchy sequences, but not all Cauchy sequences are convergent.

Can a sequence be Cauchy but not convergent?

Yes, a sequence can be Cauchy but not convergent. This can happen in cases where the sequence has a "hole" or jump, meaning there is a missing value or a value that is not part of the sequence. In these cases, the sequence can still be Cauchy because the terms are getting closer together, but it does not have a limit and therefore is not convergent.

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