Limit of a function as n approaches infinity

[agnimusayoti]

Homework Statement

Find $\lim_{n\to\infty} \frac{(-1)^n n^2}{(n+1)^2}$

Relevant Equations

Partial decomposition:
$$\frac{1}{(n+1)^2}=A/(n+1)+B/(n+1)^2)$$

If there is no $(-1)^2$ factor, I can find the limit. But, now I have no idea how to find limit for the $(-1)^\infty$. I thought $(-1)^\infty$ is an indeterminate form. So, how to modify this? Thanks!

 

[PeroK]

Homework Statement:: Find $\lim_{n\to\infty} \frac{(-1)^n n^2}{(n+1)^2}$
Relevant Equations:: Partial decomposition:
$$\frac{1}{(n+1)^2}=A/(n+1)+B/(n+1)^2)$$

If there is no $(-1)^2$ factor, I can find the limit. But, now I have no idea how to find limit for the $(-1)^\infty$. I thought $(-1)^\infty$ is an indeterminate form. So, how to modify this? Thanks!

You need to make a rule: never put $\infty$ in as a number. Something like $(-1)^{\infty}$ is horrible! You need to cut this out.

What are the options for the limit of a sequence?

 

[agnimusayoti]

How to cut $(-1)^\infty$?
Actually, it is not clear enough what do you mean by options for the limit of a sequence. As long as I know, I can expand the sequence and see if the sequence tends to a number or not. Or do I have to compute the limit?

 

[PeroK]

How to cut $(-1)^\infty$?
Actually, it is not clear enough what do you mean by options for the limit of a sequence. As long as I know, I can expand the sequence and see if the sequence tends to a number or not. Or do I have to compute the limit?

I would never write $(-1)^{\infty}$. It's just fundamentally wrong, IMHO. Instead, what can you say about:
$$\lim_{n \rightarrow \infty} (-1)^n$$

 

[agnimusayoti]

oh, I can assume $(-1)^n=x \rightarrow \ln (x)=n \ln(-1)$ which is error. Therefore the limit doesn't exist. Is that true? Thankss

 

[PeroK]

oh, I can assume $(-1)^n=x \rightarrow \ln (x)=n \ln(-1)$ which is error. Therefore the limit doesn't exist. Is that true? Thankss

Sorry, that is not mathematics. You can only take the logarithm of a positive sequence.

The first step is to prove that $\lim_{n \rightarrow \infty} (-1)^n$ does not exist. You need to use proper algebra and logic.

 

[ehild]

oh, I can assume $(-1)^n=x \rightarrow \ln (x)=n \ln(-1)$ which is error. Therefore the limit doesn't exist. Is that true? Thankss

What does the nth power of a number mean? for example, 2³=2*2*2. What is (-1)¹? What is (-1)²? What is (-1)³?... and so on.

 

[agnimusayoti]

Sorry, that is not mathematics. You can only take the logarithm of a positive sequence.

The first step is to prove that $\lim_{n \rightarrow \infty} (-1)^n$ does not exist. You need to use proper algebra and logic.

If n is an integer, then $(-1)^n$ can be graph

So, if f(n) as n is approaching infinity from left will different with f(n) as n approaching from right.

Is it can be accepted as a Mathematical prove? Because I am not used to solve proving problem

 

[agnimusayoti]

If n is an integer, then $(-1)^n$ can be graphView attachment 260958
So, if f(n) as n is approaching infinity from left will different with f(n) as n approaching from right. Therefore, $lim_{n\to\infty} (-1)^n$ does not exist

Is it can be accepted as a Mathematical prove? Because I am not used to solve proving problem.

 

[agnimusayoti]

What does the nth power of a number mean? for example, 2³=2*2*2. What is (-1)¹? What is (-1)²? What is (-1)³?... and so on.

So the idea is odd terms Will be different with even terms, right?

 

[PeroK]

Here are some things you must know.

Any sequence can do one of four things: 1) converge to some finite number; 2) diverge to to $+\infty$; 3) diverge to $-\infty$; 4) have no well-defined limit.

If the sequence is bounded (you should know what that means), then only options 1) and 4) are possible.

The sequence in your original post, $(-1)^n \frac{n^2}{(n+1)^2}$ is bounded so the only options are 1) and 4). Either it converges to a finite number, or the limit does not exist.

So, you have to find the limit; or, show that the limit does not exist.

Have you ever done a mathematical proof before?

 

[agnimusayoti]

Here are some things you must know.

Any sequence can do one of four things: 1) converge to some finite number; 2) diverge to to $+\infty$; 3) diverge to $-\infty$; 4) have no well-defined limit.

If the sequence is bounded (you should know what that means), then only options 1) and 4) are possible.

The sequence in your original post, $(-1)^n \frac{n^2}{(n+1)^2}$ is bounded so the only options are 1) and 4). Either it converges to a finite number, or the limit does not exist.

So, you have to find the limit; or, show that the limit does not exist.

Have you ever done a mathematical proof before?

Well OK, it's the first time I know bounded series. I have read again about classification series and sequences: increasing, decreasing, monotonic, bounded below, bounded above, and bounded. It makes sense if the bounded series only have 2 option, either convergent or the limit as n approaching infinity does not exist.

Yup. Usually I skip mathematical proof. But, now I will try to read again how to prove whether the limit of a function exists or does not exist.

 

[agnimusayoti]

So in the preliminary test, my task is 2: find the limit or proof whether the limit exists or does not exist, right? WUah, it's pretty difficult to do Mathematical proof

 

[PeroK]

So in the preliminary test, my task is 2: find the limit or proof whether the limit exists or does not exist, right? WUah, it's pretty difficult to do Mathematical proof

What do you think the answer is and why?

 

[ehild]

So the idea is odd terms Will be different with even terms, right?

Yes. and what is (-1)ⁿ if n is even. and what is it when n is odd?

 

[Mark44]

Homework Statement:: Find $\lim_{n\to\infty} \frac{(-1)^n n^2}{(n+1)^2}$
Relevant Equations:: Partial decomposition:
$$\frac{1}{(n+1)^2}=A/(n+1)+B/(n+1)^2)$$

I don't think this has been mentioned so far...
Partial fractions decomposition is no help in this problem. The limit in your homework statement can be found directly.

You need to make a rule: never put $\infty$ in as a number. Something like $(−1)^{\infty}$ is horrible! You need to cut this out.

@agnimusayoti, this has been mentioned before, in another thread.

 

[agnimusayoti]

Yeah I know how to find the limit without the $(-1)^n$. Like this:
$$\lim_{n\to\infty} \frac{n^2}{(n+1)^2}=\lim_{n\to\infty} \frac{1}{1+\frac{2}{n}+\frac{1}{n^2}}=1$$

In another thread, someone mention that the sequence is $a_n=\frac{n^2}{(n+1)^2}$ or the positive sequence. Then, by the preliminary test to the positive terms, the series is divergent because $\lim_{n\to\infty} \frac{n^2}{(n+1)^2} \neq 0$

Meanwhile, I can't proof $\lim_{n\to\infty} (-1)^n$ does not exist.

According to limit's definition:
$$\lim_{x\to c} f(x) =L$$
For each given $\varepsilon > 0$ there is corresponding $\delta > 0$ such that $|f(x)-L|<\varepsilon$ by $0<|x-c|<\delta$

Say, $\lim_{n\to 2} 3n=6$
$$|3n-6|<\varepsilon$$
$$3|n-2|<\varepsilon \Rightarrow |n-2|<\frac{\varepsilon}{3}$$
So, $\delta=\varepsilon/3$. Because there is corresponding $\delta$ to $\varepsilon$ then I can conclude that $\lim_{n\to 2} 3n=6$

Back to the my problem, $f(n)=(-1)^n$. Let's say $\lim{n\to\infty} f(n) = L$
There is no L, so L does not exist. This should be proofed, right?
So, I tried follow the definition
$$|(-1)^n - L|<\varepsilon$$
But,$0<|n-\infty|<\delta$
I stuck at this stage.

Is my idea was correct or I should try use another method to proof that $\lim_{n\to\infty} (-1)^n$ does not exist.

 

[Mark44]

Yeah I know how to find the limit without the $(-1)^n$. Like this:
$$\lim_{n\to\infty} \frac{n^2}{(n+1)^2}=\lim_{n\to\infty} \frac{1}{1+\frac{2}{n}+\frac{1}{n^2}}=1$$

In another thread, someone mention that the sequence is $a_n=\frac{n^2}{(n+1)^2}$ or the positive sequence. Then, by the preliminary test to the positive terms, the series is divergent because $\lim_{n\to\infty} \frac{n^2}{(n+1)^2} \neq 0$

Meanwhile, I can't proof $\lim_{n\to\infty} (-1)^n$ does not exist.

According to limit's definition:
$$\lim_{x\to c} f(x) =L$$
For each given $\varepsilon > 0$ there is corresponding $\delta > 0$ such that $|f(x)-L|<\varepsilon$ by $0<|x-c|<\delta$

Say, $\lim_{n\to 2} 3n=6$
$$|3n-6|<\varepsilon$$
$$3|n-2|<\varepsilon \Rightarrow |n-2|<\frac{\varepsilon}{3}$$
So, $\delta=\varepsilon/3$. Because there is corresponding $\delta$ to $\varepsilon$ then I can conclude that $\lim_{n\to 2} 3n=6$

Back to the my problem, $f(n)=(-1)^n$. Let's say $\lim{n\to\infty} f(n) = L$
There is no L, so L does not exist. This should be proofed, right?

Should be proven, or proved, not proofed. But you only need to do this once, and it's very easy. If there is a number L such that $\lim_{n \to \infty} (-1)^n = L$ then for any $\epsilon > 0$ that someone chooses, there is a number N such that for all $n \ge N$, $|(-1)^n - L| < \epsilon$.

Before continuing, look at the sequence $\{(-1)^n\}$, which is {1, -1, 1, -1, …}.
Take $\epsilon = 1/2$. Notice that no matter how far out in the sequence you look, any two adjacent terms are 2 units apart. There is no number L that is within 1/2 unit of all of the terms in the sequence past whatever index you want, so this sequence has no limit. It fails to converge due to oscillation.

So, I tried follow the definition
$$|(-1)^n - L|<\varepsilon$$
But,$0<|n-\infty|<\delta$
I stuck at this stage.

In part, because it makes no sense. Please stop using $\infty$ in arithmetic expressions. Furthermore, the formal definition of a limit of a sequence does not use $\delta$.
See this Wikipedia article -- https://en.wikipedia.org/wiki/Limit_of_a_sequence#Formal_definition

Is my idea was correct or I should try use another method to proof that $\lim_{n\to\infty} (-1)^n$ does not exist.

 

[agnimusayoti]

1. Hmm, the series is convergent if $\lim_{n\to \infty} S_n=S$ or the sum of series as n approaching infinity tends to finite number; not based on the terms. But the wikipedia link say about the terms. Anyway thanks for the information about formal definition in limit.(and forgive my poor English skill)
2. Is mathematical proof same with take some example and plug into the theorems and see if the theorem valid or not? Because what you do is like pick arbitrary example and give argumentation on it. Or, is it a method to proof? CMIIW

Thankss.

 

[ehild]

Meanwhile, I can't proof $\lim_{n\to\infty} (-1)^n$ does not exist.

write out some terms of the sequence $(-1)^n$: -1, 1, -1,. 1,. -1, 1 ... and so on.
Does it converge?

 

[agnimusayoti]

write out some terms of the sequence $(-1)^n$: -1, 1, -1,. 1,. -1, 1 ... and so on.
Does it converge?

No. Hmm. Looks so simple. Is it can be used as Mathematical proof by showing the some terms of the sequence? Thankss

 

[ehild]

write out some terms of the sequence $(-1)^n$: -1, 1, -1,. 1,. -1, 1 ... and so on.
Does it converge?

Yes. You can not find an N so that the next term is closer than an arbitrary small epsilon to the previous one.

 

[agnimusayoti]

No. Hmm. Looks so simple. Is it can be used as Mathematical proof by showing the some terms of the sequence? Thankss

But I don't think so. Because if I apply this to another problems, it is not clear whether the sequences tends to finite number or not. The series is $\lim_{n\to \infty} \frac{(-1)^n n}{\sqrt{n^3+1}}$ the series is -0,71+0,89-0,95+... looks like whether tends to -1 or 1. Hmm.

 

[ehild]

But I don't think so. Because if I apply this to another problems, it is not clear whether the sequences tends to finite number or not. The series is $\lim_{n\to \infty} \frac{(-1)^n n}{\sqrt{n^3+1}}$ the series is -0,71+0,89-0,95+... looks like whether tends to -1 or 1. Hmm.

If the sequence consists of two subsequences, with different limits, then the sequence is not convergent. If you assume that the limit is 1, and you choose epsilon= 0.1, the fourth term is 0.97, closer to 1 than 0.1. but the fifth term is -0.98, far away.

 

[PeroK]

No. Hmm. Looks so simple. Is it can be used as Mathematical proof by showing the some terms of the sequence? Thankss

There are at least three ways to prove that the sequence $(-1)^n$ has no limit.

1) Directly using the epsilon method, show that it cannot converge to any real number $L$.

2) Any convergent sequence is a Cauchy sequence. If you show that $(-1)^n$ is not a Cauchy sequence, then it can't be convergent.

3) If $a_n \rightarrow L$, then $a_n^2 \rightarrow L^2$. That shows that the only possible limits for $(-1)^n$ are $\pm 1$. Then you can show that neither of these is the limit using the epsilon method.

You need to familiarise yourself with all these ideas and techniques.

 

[agnimusayoti]

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.

 

[PeroK]

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 $-1$?

 

[agnimusayoti]

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

 

[agnimusayoti]

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.

 

[PeroK]

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?

 

[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.