Limit of a function as n approaches infinity

  • #26
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I try to prove.
The sequence ##{-(1)^n}_1^\infty## is -1, 1, -1, 1, -1, ...
Since the limit of the sequence can't found by inspection, so I suspect that as n approaching infinity, the limit of this sequence is 1 or -1 (by method number 3).
First, if
$$\lim_{n\to \infty} (-1)^n = 1$$
then for any small positive number ##\epsilon## there is natural number ##N## that makes ##|(-1)^n -1|<\epsilon##. If there is no natural number N, then 1 is not the limit of this sequence as n approaching infinity.
Proof:
$$|(-1)^n - 1|<\epsilon$$
$$(-1)^n <\epsilon + 1$$
Because ##\epsilon## is positive number, so ##\epsilon + 1## also a positive number (must bigger than zero). Is there any n satisfy this condition? Only the even number n that satisfy this condition. So, 1 is not the limit of this sequence.

Second, if $$\lim_{n\to \infty} (-1)^n = -1$$
then for any small positive number ##\epsilon## there is natural number ##N## that makes ##|(-1)^n + 1|<\epsilon##. If there is no natural number N, then -1 is not the limit of this sequence as n approaching infinity.
Proof:
$$|(-1)^n + 1|<\epsilon$$
$$(-1)^n <\epsilon - 1$$
Because ##\epsilon## is positive number, so ##\epsilon - 1## is a negative number (smaller than zero). Is there any n satisfy this condition? Only the odd number n that satisfy this condition. So, -1 is not the limit of this sequence.

From these 2 proofs, I can conclude that $$\lim_{n\to \infty} (-1)^n DNE$$


Now, I will try to use the more general method: number 1.
$$\lim_{n\to \infty} (-1)^n = L$$
$$|(-1)^n - L|<\epsilon$$
$$(-1)^n <\epsilon + L$$
If ##L > 0##, then only the even number of n satisfy that condition.
If ##L < 0##, then only the odd number of n satisfy that condition.
So, the limit is DNE.

Hmm. Are my works true? Thankss

Oh, I have read about Cauchy sequence, but need more explanation to identify whether a sequence is Cauchy sequence or not. May be some of you could give an example or explanation? Thanks again.
 
  • #27
PeroK
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Let me show you a proof that the limit of ##(-1)^n## is not equal to ##1##. Notice how different my proof is from yours.

Let ##\epsilon = 1##. If ##\lim_{n \rightarrow \infty} (-1)^n= 1##, then ##\exists N## such that ##n > N \ \Rightarrow \ |(-1)^n - 1| < 1##.

Note that ##|x - 1| < 1 \ \Rightarrow \ 0 < x < 2##, and in particular ##x > 0##. Therefore:

If ##\lim_{n \rightarrow \infty} = 1##, then ##\exists N## such that ##n > N \ \Rightarrow \ (-1)^n > 0##.

This does not hold, as there are always odd ##n## greater than any ##N##, with ##(-1)^n = -1 < 0##.

Hence, ##(-1)^n## does not converge to ##1##. The definition of convergence fails for ##\epsilon = 1##, for example.

Notice that, in general, what I have proved here is that:

If ##\lim_{n \rightarrow \infty} a_n = 1##, then in particular ##a_n## must eventually become positive and stay positive (##> 0##).

Notice also that I did not try to prove something for every ##\epsilon##. To prove a sequence does not converge, you must find a specific ##\epsilon## for which it fails. Note that it does not fail for ##\epsilon = 3##, for example.

If you understand this, can you now show that the limit is not ##-1##?
 
  • #28
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If limn→∞=1limn→∞=1\lim_{n \rightarrow \infty} = 1, then ∃N∃N\exists N such that n>N ⇒ (−1)n>0n>N ⇒ (−1)n>0n > N \ \Rightarrow \ (-1)^n > 0.
Actually when I tried to prove, I guessing what is different n and N. Could you help explain before I go further? Thankss
 
  • #29
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This does not hold, as there are always odd nnn greater than any NNN, with (−1)n=−1<0(−1)n=−1<0(-1)^n = -1 < 0.
Please explain this part too. Thanks again.
 
  • #30
PeroK
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Actually when I tried to prove, I guessing what is different n and N. Could you help explain before I go further? Thankss
What course are you studying? Do you have material that already explains this?
 
  • #31
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I study Mathematical Methods for Physical Sciences and uses ML Boas book. Nope, I get the notation n and N in the youtube, but the presenter do not explain what is the difference between those two number and their relevance to the proof. Apologize for my lack of knowledge.
 
  • #32
PeroK
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I study Mathematical Methods for Physical Sciences and uses ML Boas book. Nope, I get the notation n and N in the youtube, but the presenter do not explain what is the difference between those two number and their relevance to the proof. Apologize for my lack of knowledge.
Boas does not cover things like the non-convergence of the sequence ##(-1)^n##, or rigorous real analysis. I suspect you are just supposed to assume that ##(-1)^n## "obviously" does not converge to anything.

Where did you get this question?
 
  • #33
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From Boas, in Problem section 5. There are 3 problem that I failed to find the limit of the sequence because of ##(-1)^{n+1}## and ##(-1)^n## factor. I think if I can find the limit of ##(-1)^n## as n approaching infinity there will become a good experience for my learning process.

For the detail, the sequence are:
1. ##a_n=\frac{(-1)^{n+1} n^2}{n^2 + 1}##
2. ##a_n=\frac{(-1)^{n} n^2}{(n + 1)^2}##
3. ##a_n=\frac{(-1)^n n}{\sqrt{n^3 + 1}}##

But, interestingly, if I neglect the ##(-1)^n## factor, I can use the preliminary test and therefore solve the problem.
 
  • #34
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And from preliminary test I should find the limit ##\neq 0## or does not exist to conclude that sequence is divergent.
 
  • #35
PeroK
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From Boas, in Problem section 5. There are 3 problem that I failed to find the limit of the sequence because of ##(-1)^{n+1}## and ##(-1)^n## factor. I think if I can find the limit of ##(-1)^n## as n approaching infinity there will become a good experience for my learning process.

For the detail, the sequence are:
1. ##a_n=\frac{(-1)^{n+1} n^2}{n^2 + 1}##
2. ##a_n=\frac{(-1)^{n} n^2}{(n + 1)^2}##
3. ##a_n=\frac{(-1)^n n}{\sqrt{n^3 + 1}}##

But, interestingly, if I neglect the ##(-1)^n## factor, I can use the preliminary test and therefore solve the problem.
These are infinite series, not sequences! That's a completely different thing.
 
  • #36
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Gee. I thought they are similar.
 
  • #37
PeroK
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Gee. I thought they are similar.
That chapter in Boas is all about tests for convergence of infinite series. I.e.infinite sums.
 
  • #38
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That chapter in Boas is all about tests for convergence of infinite series. I.e.infinite sums.
Yeah, but for preliminary test, how can I proof those 1 and 2 series is divergent and series number 3 need further testing with the ##(-1)^n## and ##(-1)^{n+1}## factor? If the factor (-1)^n does not exist I can get the limit. And from this limit, I can conclude that series number 1 and 2 is divergent because the limit does not equal to 0. With the same idea, I can conclude that series number 3 need further testing because the limit equals to 0.
So, the main problem for me is those factor.

Since I want to focus on those factor, I assume (-1)^n is a function of n. So, I ask in this thread. So sorry if my problem irrelevance with the thread's title.
 
  • #39
PeroK
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Yeah, but for preliminary test, how can I proof those 1 and 2 series is divergent and series number 3 need further testing with the ##(-1)^n## and ##(-1)^{n+1}## factor? If the factor (-1)^n does not exist I can get the limit. And from this limit, I can conclude that series number 1 and 2 is divergent because the limit does not equal to 0. With the same idea, I can conclude that series number 3 need further testing because the limit equals to 0.
So, the main problem for me is those factor.

Since I want to focus on those factor, I assume (-1)^n is a function of n. So, I ask in this thread. So sorry if my problem irrelevance with the thread's title.
The preliminary test is whether ##a_n \rightarrow 0##. I don't know how much detail Boas goes into, but ##a_n \rightarrow 0## if and only if ##|a_n| \rightarrow 0##.

For the preliminary test, you can ignore the factor of ##(-1)^n##. Isn't that ironic?

In any case, it should be clear that this sequence does not converge to ##0##. Hence, the series fails the preliminary test.

To get through Boas, you are not required to prove these things using rigorous epsilon proofs. That's a different game altogether - perhaps an even harder game! Instead use the techniques that Boas provides.
 
  • #40
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@agnimusayoti this is why we ask for the whole question. If you'd said you were using the preliminary test for series convergence, then you would have saved a lot of time.

It's useful also to mention the book and the section. It makes life a lot easier if we know this information.
 
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  • #41
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The preliminary test is whether ##a_n \rightarrow 0##. I don't know how much detail Boas goes into, but ##a_n \rightarrow 0## if and only if ##|a_n| \rightarrow 0##.

For the preliminary test, you can ignore the factor of ##(-1)^n##. Isn't that ironic?
Huft. So far, I don't see that Boaz mention about the absolute value. No wonder one of PF member in another thread say that those test is for positive terms.
Okay, thanks for the suggestion how to ask in this forum.

Or should I learn back to sequence? I found a book that discuss infinite series from sequence. Or do you think learn from Boas is enough (with saving time consideration)?
 
  • #42
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Glad that problems are already solved. Thanksss
 
  • #43
PeroK
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Huft. So far, I don't see that Boaz mention about the absolute value. No wonder one of PF member in another thread say that those test is for positive terms.
Okay, thanks for the suggestion how to ask in this forum.

Or should I learn back to sequence? I found a book that discuss infinite series from sequence. Or do you think learn from Boas is enough (with saving time consideration)?
If you are studying mathematical methods as a physics or science major, then go with Boas. You don't want to get sidetracked into pure maths. Trust me on that.

Boas is an advanced book. There's everything in there up to Tensor analysis and Bessel functions. You will need to develop a lot of mathematical savoir faire. In chapter 1 she is moving quickly through material that should be relatively easy.
 
  • #44
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Well thanks for your reminder. I Will continue the study then . Thanks!
 
  • #45
Stephen Tashi
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Now, I will try to use the more general method: number 1.
$$\lim_{n\to \infty} (-1)^n = L$$
$$|(-1)^n - L|<\epsilon$$
$$(-1)^n <\epsilon + L$$
If ##L > 0##, then only the even number of n satisfy that condition.
If ##L < 0##, then only the odd number of n satisfy that condition.
So, the limit is DNE.

Hmm. Are my works true? Thankss
You have the correct idea and you did well for someone who isn't experienced with proofs. To improve your argument, you should state it as a proof by contradiction. As others suggest, the text you are using doesn't expect formal proofs.

You may be the sort of person who wishes to understand the details of mathematics. If so, you should study a book on Logic and understand some of the formal rules of working with the quantifiers ##\forall## and ##\exists##. I think you would find the formal study of logic easy compared to mathematical physics.
 
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