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

[tarheelborn]

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!

 

[hgfalling]

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.

 

[tarheelborn]

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?

 

[hgfalling]

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".

 

[estro]

You should show:

\exists\ \epsilon = (put\ here\ the\ right\ value)\ so\ \forall\ N\ there\ are\ m&gt;n&gt;N\ so\ that:\ |\sum_{k=n+1}^{m}a_k|\geq\epsilon

The most important thing is to understand and "feel" definitions and their meaning. You can't progress further before fully grasping them.

...
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

 

[tarheelborn]

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.