Intro Analysis: Proof that a limit = 0

[Abraham]

This is an intro to analysis problem.

I have already finished this proof (see attachment). I would like someone to check it for me. Its really short and easy. Thanks! -Abraham

Tags:
-Cauchy series
-Infinite series
-Limits

 

[Dick]

The proof is not that easy. You can't pick an N such that x_n<epsilon/n. epsilon/n isn't a constant. You didn't use that x_n is decreasing. Define a series x_n by x_n=1/n if n=2^k for some integer k and x_n=0 otherwise. lim x_n*n is not zero, but the series converges. Where does your proof fail? Now think how that can't happen if the series is decreasing.

 

[Abraham]

You can't pick an N such that x_n<epsilon/n. epsilon/n isn't a constant. You didn't use that x_n is decreasing... Where does your proof fail? Now think how that can't happen if the series is decreasing.

I'm stuck on this one. I see how the proof fails, because epsilon/n isn't a constant, but now I'm not sure how x_n decreasing helps.

The most I can claim is that n*x_n is less than n*epsilon.

I see that n diverges to infinity, and x_n converges to zero, which gives a limit of infinity*0. I don't understand this.

 

[Dick]

This is not a super easy problem. That's probably why it's 'extra credit'. You have to THINK about it. Do you know how to prove the harmonic series x_n=1/n diverges? You have to modify that proof. Do a proof by contradiction. Assume n*x_n does not approach 0. That means there is an L such that for any N there is an x_n with n>N such that n*x_n>L. Now start thinking.

 

[Abraham]

So I went about thinking, and here's my new proof. Thanks for the help!

 

[Dick]

So I went about thinking, and here's my new proof. Thanks for the help!

I don't see any contradiction there. Where is it? First you say n_i+1<n_i, then you say n_i>n_i+1. That looks like the say thing to me. The proof you want to use as a model is the most common proof the harmonic series diverges. Look at the first proof in http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

 

[Abraham]

I don't see any contradiction there. Where is it? First you say n_i+1<n_i, then you say n_i>n_i+1. That looks like the say thing to me. The proof you want to use as a model is the most common proof the harmonic series diverges. Look at the first proof in http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

Sorry, I incorrectly typed this into Latex. I meant to say that the sequence n_i is a strictly increasing sequence. It is the sequence of all n for which n*x_n is greater or equal to epsilon. Progressing along the natural numbers, n(1) must be less than n(2), and n(i) < n(i+1). The contradiction arises when the decreasing x_n sequence: x(i) > x(i+1), implies that n(i) > n(i+1).

That is, 1/n(i) > 1/n(i+1) implies n(i) < n(i+1), a contradiction.

My rationale for doing it this way instead of a modified harmonic series proof, is that I don't know any values of the n_i sequence. I feel as though I need actual numbers to show that a series is unbounded. The harmonic series proof of divergence uses the fact that we can group terms in such a way that we have a constant unbounded sequence of 1/2.

I am guaranteed the existence of n's from the statement, "there exist n such that n*x_n is greater or equal to epsilon". But I gain no numerical information to show that the series of those n is unbounded. What do you think of this?

I will upload a corrected PDF soon

 

[Dick]

Sorry, I incorrectly typed this into Latex. I meant to say that the sequence n_i is a strictly increasing sequence. It is the sequence of all n for which n*x_n is greater or equal to epsilon. Progressing along the natural numbers, n(1) must be less than n(2), and n(i) < n(i+1). The contradiction arises when the decreasing x_n sequence: x(i) > x(i+1), implies that n(i) > n(i+1).

That is, 1/n(i) > 1/n(i+1) implies n(i) < n(i+1), a contradiction.

My rationale for doing it this way instead of a modified harmonic series proof, is that don't know any values of the n_i sequence. I feel as though I need actual numbers to show that a series is unbounded. The harmonic series proof of divergence uses the fact that we can group terms in such a way that we have a constant unbounded sequence of 1/2.

I am guaranteed the existence of n's from the statement, "there exist n such that n*x_n is greater or equal to epsilon". But I gain no numerical information to show that the series of those n is unbounded. What do you think of this?

I will upload a corrected PDF soon

Do NOT upload a new PDF if you are thinking along those lines. It will be wrong. You really need to prove the sum of the series is unbounded. Here's a REALLY BIG HINT. Pick N such that x_N>e/N. Now there is a M>2N such that x_M>e/M, right? Can you suggest a lower bound for the sum of all of the elements between x_N and x_M?

 

[Abraham]

Ah, I was so convinced I could do something clever using just the ordering. Also, if you have the time to explain, why was the previous proof incorrect? I arrived at a contradiction. I see that I didn't make use of all the hypotheses; I'm wondering if that makes it an insufficient contradiction?

Anyways, many thanks for the hints.

-A.

 

[Dick]

Ah, I was so convinced I could do something clever using just the ordering. Also, if you have the time to explain, why was the previous proof incorrect? I arrived at a contradiction. I see that I didn't make use of all the hypotheses; I'm wondering if that makes it an insufficient contradiction?

Anyways, many thanks for the hints.

-A.

The first proof is wrong because there is no contradiction. I said this before. Where is it? n_i+1<n_i doesn't contradict n_i>n_i+1.

The new proof doesn't seem to have anything to do with the given series x_n. You just replaced the x_n with e/n. Why don't you stop posting completed proofs and just try to work in out in pieces. Start with the question I asked in post 8.