Proof of convergence

  • #1
Use only the definition of convergence to prove ## \dfrac{cos(n)}{n} \rightarrow 0 ##

my proof:

given ## \epsilon > 0 ## ##\exists N \in \mathbb{R} ## s.t. n>N => ## |\dfrac{cos(n)}{n}| < \epsilon ##

## |cos(n)| \leq 1 ## therfore ## \dfrac{|cos(n)|}{|n|} \leq \dfrac{1}{|n|} < \epsilon ## so ## |n| > \dfrac{1}{\epsilon} ##
so choose big N to be 1/e and we get ## |\dfrac{cos(n)}{n}| < \epsilon ##

is this proof ok?

I am also confused about this

from here: ## \dfrac{|cos(n)|}{|n|} < \epsilon ## we can do ## -\epsilon |n| < cos(n) < \epsilon |n| ## and if we consider ## cos(n) < \epsilon |n| ## we get ## \epsilon |n| > -1 ## and therefore ## n > \dfrac{-1}{\epsilon} ## and ## n < \dfrac{1}{\epsilon} ## why do I get two different results contradicting with my first proof, i'm assuming i've gone wrong so if someone could correct me where, please.
 

Answers and Replies

  • #2
similarly if I was to prove it using the sandwich theorem, would it be ok in saying ## -1 \leq cos(n) \leq 1 ## and then dividing by n? i.e. ## -1/n \leq cos(n)/n \leq 1/n ## seeing as we don't know if n is negative or not?
 
  • #3
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,557
767
Use only the definition of convergence to prove ## \dfrac{cos(n)}{n} \rightarrow 0 ##

my proof:

given ## \epsilon > 0 ## ##\exists N \in \mathbb{R} ## s.t. n>N => ## |\dfrac{cos(n)}{n}| < \epsilon ##

That is a statement of what you are to prove.

## |cos(n)| \leq 1 ## therfore ## \dfrac{|cos(n)|}{|n|} \leq \dfrac{1}{|n|} < \epsilon ## so ## |n| > \dfrac{1}{\epsilon} ##
so choose big N to be 1/e and we get ## |\dfrac{cos(n)}{n}| < \epsilon ##

is this proof ok?

You have the idea but it would be better to rewrite it in a different order:

Suppose ##\epsilon > 0##. Pick ##N## so that ##N>\frac 1 \epsilon##. Then if ##n > N##, then ##n>\frac 1 \epsilon## so$$
\frac {|\cos(n)|} {n} \le \frac 1 n <\epsilon$$That's all you need to write.
 
  • #4
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,557
767
similarly if I was to prove it using the sandwich theorem, would it be ok in saying ## -1 \leq cos(n) \leq 1 ## and then dividing by n? i.e. ## -1/n \leq cos(n)/n \leq 1/n ## seeing as we don't know if n is negative or not?

You are going to have ##\epsilon > 0## and ##N > \frac 1 \epsilon##, so your ##n## will be positive if ##n>N##.
 
  • #5
329
34
You didn't say, but from the way you did the proof can we assume n ##\rightarrow \infty##? If so, the proof is fine, altho it seems to have a typo where you wrote "e" instead of ##\epsilon##.
I am also confused about this

from here: |cos(n)||n|<ϵ we can do −ϵ|n|<cos(n)<ϵ|n| and if we consider cos(n)<ϵ|n| we get ϵ|n|>−1 and therefore n>−1ϵ and n<1ϵ why do I get two different results contradicting with my first proof, i'm assuming i've gone wrong so if someone could correct me where, please.

It's |cos(n)|/|n| < ##\epsilon##. If n##\rightarrow \infty## then n > 0, and you don't need the absolute values on n. So you have |cos(n)| < n##\epsilon## where n >1/##\epsilon##. This means only that ##\epsilon##n > 1. So all we are saying here is that |cos(n)| is less than a number which is > 1. That's perfectly fine. Where we would be in trouble is if we concluded that n##\epsilon## < 1, because then it wouldn't be true that |cos(n)| < n##\epsilon## for every n.

The computation you did somehow got the inequalities mixed up. You should have -##\epsilon##n < -1, because you are comparing with the smallest value of cos(n).

Questioning things this way is very good. It forces you to get things absolutely clear, rather than half understanding.
 
  • #6
You didn't say, but from the way you did the proof can we assume n ##\rightarrow \infty##? If so, the proof is fine, altho it seems to have a typo where you wrote "e" instead of ##\epsilon##.

It's |cos(n)|/|n| < ##\epsilon##. If n##\rightarrow \infty## then n > 0, and you don't need the absolute values on n. So you have |cos(n)| < n##\epsilon## where n >1/##\epsilon##. This means only that ##\epsilon##n > 1. So all we are saying here is that |cos(n)| is less than a number which is > 1. That's perfectly fine. Where we would be in trouble is if we concluded that n##\epsilon## < 1, because then it wouldn't be true that |cos(n)| < n##\epsilon## for every n.

The computation you did somehow got the inequalities mixed up. You should have -##\epsilon##n < -1, because you are comparing with the smallest value of cos(n).

Questioning things this way is very good. It forces you to get things absolutely clear, rather than half understanding.

continuing from ## -\epsilon|n| < cos(n) < \epsilon|n| ## and looking at the upper inequality only i.e. ## cos(n) < \epsilon|n|## and you said we take the smallest value of cos(n) which is -1 right? so ## \epsilon |n| > -1 ## so ## n > -1/\epsilon ## ? If, however, we take the largest value of cos(n) we get the correct answer. Which ones should we be taking if we had ## -\epsilon|n| < cos(n) < \epsilon|n|##
 
  • #7
You are going to have ##\epsilon > 0## and ##N > \frac 1 \epsilon##, so your ##n## will be positive if ##n>N##.

yes but assuming I did not prove that ## n > 1/\epsilon ## then how would I know it's ok to divide by n?
 
  • #8
329
34
continuing from ## -\epsilon|n| < cos(n) < \epsilon|n| ## and looking at the upper inequality only i.e. ## cos(n) < \epsilon|n|## and you said we take the smallest value of cos(n) which is -1 right? so ## \epsilon |n| > -1 ## so ## n > -1/\epsilon ## ? If, however, we take the largest value of cos(n) we get the correct answer. Which ones should we be taking if we had ## -\epsilon|n| < cos(n) < \epsilon|n|##

When I said "smallest value of the cosine" I was looking at the left hand inequality. Trying to explain further:

The cosine as you know oscillates between 1 and -1. -n##\epsilon## < cos(n) for every n requires that -n ##\epsilon \le -1##. You have chosen your n > N so that inequality is true. Similarly cos(n) < n##\epsilon## for every n would require n##\epsilon## > 1. Again, because of your choice of n that is true.

Once you've chosen an n so that |cos(n)|/n < ##\epsilon##, all other arithmetic restatements of it have to be true. The restatement you have given here is certainly true because of your choice of n vs the ##\epsilon## that was chosen. Your statement is not useful for convergence purposes, but does serve to force us to think about what these inequalities really mean, and that is a good thing.

Do you understand it better now? Or do we need more discussion?
 
  • #9
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,557
767
yes but assuming I did not prove that ## n > 1/\epsilon ## then how would I know it's ok to divide by n?

I'm not sure why this is an issue for you. You are proving a limit as ##n\to\infty##. You are really only interested in large positive ##n##.
 
  • #10
329
34
yes but assuming I did not prove that ## n > 1/\epsilon ## then how would I know it's ok to divide by n?

As ##\epsilon## gets smaller and smaller, N gets larger and larger. If you want to prove the series converges you have to choose an N, and you properly chose it so that N > 1/##\epsilon##, which means for n>N n >1/##\epsilon##. So there is certainly no worry about dividing by n.

You didn't "prove" n >1/##\epsilon##. You chose it in order to show that |cos(n)|/n goes to 0 as n ##\rightarrow \infty##. So no matter what ##\epsilon## you choose there is always an N so that for every n > N |cos(n)|/n < ##\epsilon##.

I think it might help you to review again the definition of convergence as n ##\rightarrow \infty##. I get the impression you haven't quite grasped the idea, even though you did the computation correctly.
 

Related Threads on Proof of convergence

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
776
  • Last Post
Replies
14
Views
1K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
8
Views
907
  • Last Post
Replies
13
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
5
Views
510
  • Last Post
Replies
3
Views
1K
Top