Intro Analysis: Proof that a limit = 0

In summary: You didn't use that x_n is decreasing. The proof is not that easy. You can't pick an N such that x_n<epsilon/n. epsilon/n isn't a constant. Where does your proof fail?In summary, the conversation discusses a proof for an analysis problem involving Cauchy and infinite series. The initial proof presented by Abraham is found to be incorrect as it does not take into account all the necessary hypotheses. The conversation then delves into finding a correct proof, with hints and clarifications provided by the expert. Ultimately, it is suggested to use a lower bound for the sum of all elements between two specific terms in the series in order to prove its divergence.
  • #1
Abraham
69
0
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
 

Attachments

  • Math 414 Extra Credit.pdf
    71.7 KB · Views: 256
Physics news on Phys.org
  • #2
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.
 
Last edited:
  • #3
Dick said:
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.
 
  • #4
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.
 
Last edited:
  • #5
So I went about thinking, and here's my new proof. Thanks for the help!
 

Attachments

  • Math 414 Extra Credit.pdf
    86.7 KB · Views: 275
  • #6
Abraham said:
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)
 
  • #7
Dick said:
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
 
  • #8
Abraham said:
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?
 
  • #9
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.
 

Attachments

  • Math 414 Extra Credit.pdf
    93.3 KB · Views: 211
Last edited:
  • #10
Abraham said:
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.
 

1. What is the concept of a limit in mathematics?

The limit of a function is the value that a function approaches as the input variable gets closer and closer to a specific value. It is used to describe the behavior of a function at a particular point and is an essential concept in calculus and analysis.

2. How is a limit defined in mathematical notation?

A limit is defined using the notation of lim as follows: limx→a f(x) = L, where a is the point at which the function is being evaluated, f(x) is the function, and L is the limit value.

3. What does it mean if a limit equals 0?

If a limit equals 0, it means that the function is approaching 0 as the input variable gets closer to a specific value. This indicates that the function is getting smaller and smaller, but never reaches 0 at that particular point.

4. How do you prove that a limit equals 0?

To prove that a limit equals 0, you need to show that the function value can be made arbitrarily close to 0 by choosing appropriate values for the input variable. This can be done using the epsilon-delta definition of a limit or by using algebraic manipulation and theorems of limits.

5. What are some practical applications of understanding limits?

Understanding limits is crucial in many areas of science and engineering. It is used in physics to calculate the instantaneous velocity and acceleration of an object, in economics to determine the marginal cost and revenue of a product, and in computer science to analyze algorithms and optimize their performance.

Similar threads

  • Calculus and Beyond Homework Help
Replies
12
Views
698
  • Calculus and Beyond Homework Help
Replies
3
Views
503
  • Calculus and Beyond Homework Help
Replies
4
Views
858
  • Calculus and Beyond Homework Help
Replies
3
Views
914
  • Calculus and Beyond Homework Help
Replies
19
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
764
  • Calculus and Beyond Homework Help
Replies
1
Views
624
  • Calculus and Beyond Homework Help
Replies
1
Views
688
  • Calculus and Beyond Homework Help
Replies
1
Views
455
  • Calculus and Beyond Homework Help
Replies
20
Views
1K
Back
Top