Limit of a function as n approaches infinity

[agnimusayoti]

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.

 

[PeroK]

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?

 

[agnimusayoti]

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.

 

[agnimusayoti]

And from preliminary test I should find the limit $\neq 0$ or does not exist to conclude that sequence is divergent.

 

[PeroK]

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.

 

[agnimusayoti]

Gee. I thought they are similar.

 

[PeroK]

Gee. I thought they are similar.

That chapter in Boas is all about tests for convergence of infinite series. I.e.infinite sums.

 

[agnimusayoti]

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.

 

[PeroK]

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.

 

[PeroK]

@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.

 

[agnimusayoti]

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)?

 

[agnimusayoti]

Glad that problems are already solved. Thanksss

 

[PeroK]

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.

 

[agnimusayoti]

Well thanks for your reminder. I Will continue the study then . Thanks!

 

[Stephen Tashi]

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.