Prove Sum Approximation Theorem

[Potatochip911]

This is incorrect. According to the definition of "limit", the statement means that for all $\varepsilon>0$, there's a positive integer N such that the following implication holds for all positive integers n.
$$n\geq N\ \Rightarrow\ |cx_n-cx|<\varepsilon.$$
The first "for all", the "there exists" and the second "for all" are all essential parts of the statement. If you miss anyone of them, the proof will fail.

Also note that it wouldn't make much sense for the definition of limit to use the notation "lim" in the explanation of what a limit is. The definition specifies what it means to say that a number is a limit of a sequence. Once we have done that, and proved that no sequence has more than one limit, we can define the notation "lim" by saying that $\lim_n x_n$ denotes the unique limit of the sequence $(x_n)_{n=1}^\infty$.That's not an assumption. It's what you're trying to prove. So you absolutely cannnot use it here.

If you need to use it in some other proof, it wouldn't look like that. You would have to make it clear how you're using the "for all $\varepsilon>0$" (by saying "let $\varepsilon>0$"), how you're using the "there exists" (to make a choice of N that will be useful later) and how you're using the "for all n" (by making a "for all n" statement). It would be very similar to what I did in the first sentence of the proof in post #53. The statement would involve $|cx-cx_n|$, not $|cx-c\lim_n x_n|$. It wouldn't involve the "lim" notation at all, since it's a statement made by a definition that logically had to be stated before that notation was introduced.

Let $\varepsilon>0$, Let $x=\lim_{n}x_{n}$, let $n$ be a positive integer such that the following implication holds for all positive integers $N$. $$n\geq N \Rightarrow |x_{n}-x|<\frac{\varepsilon}{|c|} \\
|cx_{n}-cx|<\varepsilon$$ So is this what I wanted to prove?

 

[Fredrik]

Let $\varepsilon>0$, Let $x=\lim_{n}x_{n}$, let $n$ be a positive integer such that the following implication holds for all positive integers $N$. $$n\geq N \Rightarrow |x_{n}-x|<\frac{\varepsilon}{|c|} \\
|cx_{n}-cx|<\varepsilon$$ So is this what I wanted to prove?

By letting ε be an arbitrary positive real number, and by choosing N this way, you have covered the first "for all" and the "there exists" in the statement that you want to prove. To finish the proof, you need to show that the second "for all" statement holds for this particular ε and this particular N. So the next thing you say should be "let n be a positive integer such that $n\geq N$". Then you prove that last inequality.

 

[Potatochip911]

After rereading it I think it would make more sense if I changed it to "Let N be a positive integer such that the following implication holds for all positive integers n" since I think there is only one N but there are more integers larger than or equal to N, "n's", that will satisfy the limit. And by the second "for all" statement you are referring to $|x_{n}-x|<\frac{\varepsilon}{|c|}$?

 

[Fredrik]

Yes, I agree with what you're saying about N and n. What I called the second "for all" statement is the "for all positive integers n" statement.

 

[Potatochip911]

Yes, I agree with what you're saying about N and n. What I called the second "for all" statement is the "for all positive integers n" statement.

Is this correct then? Let $x=\lim_{n}x_{n}$, Let $\varepsilon>0$ and Let N be a positive integer such that the following implication holds for all positive integers n: $$n \geq N \Rightarrow |x_{n}-x|<\frac{\varepsilon}{|c|} \\
|cx_{n}-cx|<\varepsilon
$$

 

[Fredrik]

Is this correct then? Let $x=\lim_{n}x_{n}$, Let $\varepsilon>0$ and Let N be a positive integer such that the following implication holds for all positive integers n: $$n \geq N \Rightarrow |x_{n}-x|<\frac{\varepsilon}{|c|} \\
|cx_{n}-cx|<\varepsilon
$$

Yes, that's a good start, but the last line is unnecessary. Also, it's weird to write that inequality on the line below the implication, because when you put statements on separate lines, each line should be implied by the one before it.

 

[Potatochip911]

Yes, that's a good start, but the last line is unnecessary. Also, it's weird to write that inequality on the line below the implication, because when you put statements on separate lines, each line should be implied by the one before it.

So to imply that I should multiply both sides by $|c|$ on the line above?

 

[Fredrik]

So to imply that I should multiply both sides by $|c|$ on the line above?

The first line isn't an inequality that you can multiply by |c| to get the last line. It's an implication. The implication can be true even if the inequality is false. So the truth of the implication doesn't guarantee that the new inequality holds. Further, the implication isn't a standalone statement. It's just the ending of the definition of N.

What you have done so far is to use the assumption to assign a value to N that will help you prove that $\lim_n (cx_n)=c\lim_nx_n$. So now you need to go ahead and prove that. To do this, it's essential that you understand what the statement $\lim_n (cx_n)=c\lim_nx_n$ means.

Edit: The first thing you should do is to write down (on a piece of paper or in a forum post, but not in the actual proof) the "for all ε" statement that according to the definition of "limit" is equivalent to $\lim_n (cx_n)=c\lim_nx_n$. It should be easier to prove the statement when you have it in front of you.

That statement is of the type "for all ε there exists N such that for all n,..." In the part of the proof that you have already typed up, you have already let ε be an arbitrary positive real number. So you can focus on the rest of the statement, "there exists N such that for all n,..." But you have also already assigned a value to N, so the only thing left to do is to prove that that N is such that the "for all n" statement holds.

 

[Potatochip911]

The first line isn't an inequality that you can multiply by |c| to get the last line. It's an implication. The implication can be true even if the inequality is false. So the truth of the implication doesn't guarantee that the new inequality holds. Further, the implication isn't a standalone statement. It's just the ending of the definition of N.

What you have done so far is to use the assumption to assign a value to N that will help you prove that $\lim_n (cx_n)=c\lim_nx_n$. So now you need to go ahead and prove that. To do this, it's essential that you understand what the statement $\lim_n (cx_n)=c\lim_nx_n$ means.

Edit: The first thing you should do is to write down (on a piece of paper or in a forum post, but not in the actual proof) the "for all ε" statement that according to the definition of "limit" is equivalent to $\lim_n (cx_n)=c\lim_nx_n$. It should be easier to prove the statement when you have it in front of you.

That statement is of the type "for all ε there exists N such that for all n,..." In the part of the proof that you have already typed up, you have already let ε be an arbitrary positive real number. So you can focus on the rest of the statement, "there exists N such that for all n,..." But you have also already assigned a value to N, so the only thing left to do is to prove that that N is such that the "for all n" statement holds.

For $\varepsilon>0$ there exists a positive integer $N$ such that the following holds for all positive integers $n$ $$
n\geq N \Rightarrow |\lim_{n}(cx_{n})-c\lim_{n}(x_{n})|<\varepsilon
$$
Is this the inequality that I want?

 

[Fredrik]

For $\varepsilon>0$ there exists a positive integer $N$ such that the following holds for all positive integers $n$ $$
n\geq N \Rightarrow |\lim_{n}(cx_{n})-c\lim_{n}(x_{n})|<\varepsilon
$$
Is this the inequality that I want?

The statement is structured correctly, with the correct "for all"s and "there exists", but the inequality is not the right one. Keep in mind that the logical order of things is this:

1. Define what is meant by saying that a real number is a limit of a sequence.
2. Prove that no sequence has more than one limit.
3. Introduce the notation $\lim_n x_n$ for the unique limit of the sequence $(x_n)_{n=1}^\infty$.

Because of this, it wouldn't make sense for the "lim" notation (introduced in step 3) to show up in a straightforward application of the definition of "limit" (step 1). Also, if you look at post #18 again, you will see that the "lim" notation doesn't appear in my definition of "limit".

 

[Potatochip911]

The statement is structured correctly, with the correct "for all"s and "there exists", but the inequality is not the right one. Keep in mind that the logical order of things is this:

1. Define what is meant by saying that a real number is a limit of a sequence.
2. Prove that no sequence has more than one limit.
3. Introduce the notation $\lim_n x_n$ for the unique limit of the sequence $(x_n)_{n=1}^\infty$.

Because of this, it wouldn't make sense for the "lim" notation (introduced in step 3) to show up in a straightforward application of the definition of "limit" (step 1). Also, if you look at post #18 again, you will see that the "lim" notation doesn't appear in my definition of "limit".

So for #1: If c is a real number, Let $\varepsilon>0$, Let N be a positive integer so that the following implication holds for all positive integers n $$
n\geq N \Rightarrow |x_{n}-c|<\varepsilon $$
I am not really sure where to go with #2 other than that I have: Let $\varepsilon>0$, Let N be a positive integer so that the following implication holds for all positive integers n $$
n\geq N \Rightarrow |x_{n}-a|<\varepsilon$$ and $$n\geq N \Rightarrow |x_{n}-b|<\varepsilon$$ where $a$ and $b$ are 2 different limits of the sequence. I am having difficulty thinking of a way to show that this is false.

 

[Fredrik]

I guess I didn't express myself clearly enough. I was just trying to explain why the "lim" notation can't be a part of what the definition of "limit" says that a limit is. The notation doesn't make sense until you have proved that no sequence has more than one limit, and you can't prove that without using the definition. To someone who understands this, it's obvious that the answer to the question of what $\lim_n (cx_n)=cx$ means can't involve the "lim" notation. I felt that I had to explain this, since your answer to that question in post #69 was referring to limits inside the "for all ε" statement.

I wasn't asking you to do 1,2,3 now. Number 1 is what I did in post #18. Number 2 is the first theorem I asked you to prove in #54. (It's more difficult than the proof you're working on now). Number 3 is just the decision to use the "lim" notation to denote the unique limit of a sequence that has a limit.

So for #1: If c is a real number, Let $\varepsilon>0$, Let N be a positive integer so that the following implication holds for all positive integers n $$
n\geq N \Rightarrow |x_{n}-c|<\varepsilon $$

If you meant that this is what it means to say that c is a limit of $(x_n)_{n=1}^\infty$, then yes, this is the way to do it.

I am not really sure where to go with #2 other than that I have: Let $\varepsilon>0$, Let N be a positive integer so that the following implication holds for all positive integers n $$
n\geq N \Rightarrow |x_{n}-a|<\varepsilon$$ and $$n\geq N \Rightarrow |x_{n}-b|<\varepsilon$$ where $a$ and $b$ are 2 different limits of the sequence. I am having difficulty thinking of a way to show that this is false.

This is a pretty good start of the proof of the result that no sequence has more than one limit. It doesn't quite work, but it's not possible to see that yet. The key part of the proof is the following application of the triangle inequality: $|a-b|=|a-x_N+x_N-b|\leq |a-x_N|+|x_N-b|$. At this point, you will want N to be such that the sum on the right is less than ε. This will happen for example if N is such that each of the terms is less than ε/2.

I would however like you to finish the proof of the other theorem first. It's probably the easiest of all proofs that involve the definition of "limit". You have correctly written down the statement that assigns a value to N. To finish the proof, you have to use that choice to prove that $\lim_n (cx_n)=cx$. This should be easy if you're able to state the "for all ε" statement that by definition of "limit" and the "lim" notation is equivalent to $\lim_n (cx_n)=cx$.

 

[Potatochip911]

I guess I didn't express myself clearly enough. I was just trying to explain why the "lim" notation can't be a part of what the definition of "limit" says that a limit is. The notation doesn't make sense until you have proved that no sequence has more than one limit, and you can't prove that without using the definition. To someone who understands this, it's obvious that the answer to the question of what $\lim_n (cx_n)=cx$ means can't involve the "lim" notation. I felt that I had to explain this, since your answer to that question in post #69 was referring to limits inside the "for all ε" statement.

I wasn't asking you to do 1,2,3 now. Number 1 is what I did in post #18. Number 2 is the first theorem I asked you to prove in #54. (It's more difficult than the proof you're working on now). Number 3 is just the decision to use the "lim" notation to denote the unique limit of a sequence that has a limit.

If you meant that this is what it means to say that c is a limit of $(x_n)_{n=1}^\infty$, then yes, this is the way to do it.This is a pretty good start of the proof of the result that no sequence has more than one limit. It doesn't quite work, but it's not possible to see that yet. The key part of the proof is the following application of the triangle inequality: $|a-b|=|a-x_N+x_N-b|\leq |a-x_N|+|x_N-b|$. At this point, you will want N to be such that the sum on the right is less than ε. This will happen for example if N is such that each of the terms is less than ε/2.

I would however like you to finish the proof of the other theorem first. It's probably the easiest of all proofs that involve the definition of "limit". You have correctly written down the statement that assigns a value to N. To finish the proof, you have to use that choice to prove that $\lim_n (cx_n)=cx$. This should be easy if you're able to state the "for all ε" statement that by definition of "limit" and the "lim" notation is equivalent to $\lim_n (cx_n)=cx$.

So I want to show that $$|cx-\lim_{n}(cx_{n})|<\varepsilon$$?

 

[Fredrik]

So I want to show that $$|cx-\lim_{n}(cx_{n})|<\varepsilon$$?

No, that's not it. You keep putting a notation for "the unique limit of..." in these statements. I pointed out that this is a mistake in posts 46, 48, 58, 60, 70 and 72. In the last three of those, I also explained why. I will try again, with a less mathematical example.

Definition: A goose is said to be a gander if it's an adult male.
Question: If it's known that Kevin is a goose, what does the statement "Kevin is a gander" mean?
Correct answer: "Kevin is an adult male".
Incorrect answer (example): "Kevin is a bag of potato chips"
Nonsense answers (examples): "Kevin is an adult gander", "Kevin is a male gander", "geese are ganders", "Lisa is a gander",...

Your answers to the question of what $\lim_n(cx_n)=cx$ means are so far all of the same type as the answers in the last line above. You're supposed to use the definition of limit to answer the question. That definition isn't making references to the limit concept in the explanation of what that concept means. It couldn't possibly do that. This would be like a definition of "gander" that says that a gander is a male gander.

 

[Potatochip911]

No, that's not it. You keep putting a notation for "the unique limit of..." in these statements. I pointed out that this is a mistake in posts 46, 48, 58, 60, 70 and 72. In the last three of those, I also explained why. I will try again, with a less mathematical example.

Definition: A goose is said to be a gander if it's an adult male.
Question: If it's known that Kevin is a goose, what does the statement "Kevin is a gander" mean?
Correct answer: "Kevin is an adult male".
Incorrect answer (example): "Kevin is a bag of potato chips"
Nonsense answers (examples): "Kevin is an adult gander", "Kevin is a male gander", "geese are ganders", "Lisa is a gander",...

Your answers to the question of what $\lim_n(cx_n)=cx$ means are so far all of the same type as the answers in the last line above. You're supposed to use the definition of limit to answer the question. That definition isn't making references to the limit concept in the explanation of what that concept means. It couldn't possibly do that. This would be like a definition of "gander" that says that a gander is a male gander.

I am really confused by this limit stuff. Your explanation about gander makes sense but I'm completely lost as to how it relates to the mistake I'm making. Also what is the notation for "the unique limit of"? Is that what $|f(x)-L|<\varepsilon$ is? Is one of the mistakes I'm making that I keep using an actual limit inside $|f(x)-L|<\varepsilon$ instead of just a number for the limit? In your example the limit is 0 so you ended up with $|\frac{1}{n}-0|<\varepsilon$. This makes sense to me but I don't understand it very well when it's of the form $\lim_{n}(cx_{n})=cx_{n}$

 

[Fredrik]

I am really confused by this limit stuff. Your explanation about gander makes sense but I'm completely lost as to how it relates to the mistake I'm making.

The answer to the question of what it means to say that Kevin is a gander can't possibly involve the word "gander". The answer to the question of what it means to say that $cx$ is a limit of the sequence $(cx_n)_{n=1}^\infty$ can't possibly involve a notation that means exactly "the unique limit of the sequence $(cx_n)_{n=1}^\infty$". You're using a notation for "the limit of this sequence" inside the statement that's supposed to say what the limit of the sequence is.

Also what is the notation for "the unique limit of"?

I mean notations like $\lim_n$ and $\sum_{n=1}^\infty$. The unique limit of the sequence $(x_n)_{n=1}^\infty$ is denoted by $\lim_n x_n$. The unique limit of the sequence of partial sums of the series $\sum_{n=1}^\infty a_n$ is denoted by $\sum_{n=1}^\infty a_n$. (Yes, in the latter case, the same string of text represents both the sequence of partial sums and its limit).

Is one of the mistakes I'm making that I keep using an actual limit inside $|f(x)-L|<\varepsilon$ instead of just a number for the limit?

Yes. You can put a symbol like L there, but you certainly can't put $\lim_{x\to a} f(x)$ inside the statement that says what the limit of $f$ at $a$ is.

You're supposed to write an expression (=string of text) that represents the real number that we want to say is the limit of the sequence. What you've been doing instead is to write an expression that means "the limit of this sequence". That doesn't make sense inside the statement that's supposed to say what the limit of the sequence is.

In your example the limit is 0 so you ended up with $|\frac{1}{n}-0|<\varepsilon$. This makes sense to me but I don't understand it very well when it's of the form $\lim_{n}(cx_{n})=cx_{n}$

The limit of a sequence is typically not equal to the nth term. What we're dealing with is $\lim_n(cx_n)=cx$. I'm not sure what it is about this notation that you find confusing. Do you know what the terms of this sequence are? Do you know what its limit is? If you do, you can't go wrong if you just look at the definition of "limit" and identify the notation for terms of the sequence, and the notation for the real number that's said to be the limit of the sequence, and then just replace the former with the notation for terms of this sequence, and the latter with the notation for this limit.

 

[Potatochip911]

The answer to the question of what it means to say that Kevin is a gander can't possibly involve the word "gander". The answer to the question of what it means to say that $cx$ is a limit of the sequence $(cx_n)_{n=1}^\infty$ can't possibly involve a notation that means exactly "the unique limit of the sequence $(cx_n)_{n=1}^\infty$". You're using a notation for "the limit of this sequence" inside the statement that's supposed to say what the limit of the sequence is.I mean notations like $\lim_n$ and $\sum_{n=1}^\infty$. The unique limit of the sequence $(x_n)_{n=1}^\infty$ is denoted by $\lim_n x_n$. The unique limit of the sequence of partial sums of the series $\sum_{n=1}^\infty a_n$ is denoted by $\sum_{n=1}^\infty a_n$. (Yes, in the latter case, the same string of text represents both the sequence of partial sums and its limit).Yes. You can put a symbol like L there, but you certainly can't put $\lim_{x\to a} f(x)$ inside the statement that says what the limit of $f$ at $a$ is.

You're supposed to write an expression (=string of text) that represents the real number that we want to say is the limit of the sequence. What you've been doing instead is to write an expression that means "the limit of this sequence". That doesn't make sense inside the statement that's supposed to say what the limit of the sequence is.The limit of a sequence is typically not equal to the nth term. What we're dealing with is $\lim_n(cx_n)=cx$. I'm not sure what it is about this notation that you find confusing. Do you know what the terms of this sequence are? Do you know what its limit is? If you do, you can't go wrong if you just look at the definition of "limit" and identify the notation for terms of the sequence, and the notation for the real number that's said to be the limit of the sequence, and then just replace the former with the notation for terms of this sequence, and the latter with the notation for this limit.

So the unique limit of a sequence is just referring to the last term of the sequence?

 

[Fredrik]

So the unique limit of a sequence is just referring to the last term of the sequence?

Absolutely not. These sequences don't have a last term. What would be the last term of $\left(\frac 1 n\right)_{n=1}^\infty$?

 

[Potatochip911]

Absolutely not. These sequences don't have a last term. What would be the last term of $\left(\frac 1 n\right)_{n=1}^\infty$?

Okay so they don't really having a last term because they're infinitely long. When we use the notation $\lim_{n}x_{n}$ are we referring to the limit of the sequence $(x_{n})_{n=1}^{\infty}$ or the the limit of the n^th term?

 

[Fredrik]

Okay so they don't really having a last term because they're infinitely long. When we use the notation $\lim_{n}x_{n}$ are we referring to the limit of the sequence $(x_{n})_{n=1}^{\infty}$ or the the limit of the n^th term?

It's the limit of the sequence. The terms are symbols that represent real numbers. Symbols don't have limits. Real numbers don't have limits.

 

[bhobba]

Okay so they don't really having a last term because they're infinitely long. When we use the notation $\lim_{n}x_{n}$ are we referring to the limit of the sequence $(x_{n})_{n=1}^{\infty}$ or the the limit of the n^th term?

Colloquially this is what is known as doing your epsilonics.

Its part of the area called analysis which is basically a very careful development of the ideas of calculus - and a considerable extension of it.

Basic to it is a careful definition of what a limit is. I could give you the definition, but to understand it you really need to see examples and why it has that definition.

Check out:
http://www.jirka.org/ra/realanal.pdf

Once that is understood then sorting out your issues will likely be easy.

BTW don't be worried if you are finding this hard. Most do. When I did my degree there were a number of people on their third attempt to pass their introductory analysis course. Just before I graduated they changed the course so it was integrated with other calculus subjects, the hope being students will find that easier.

Thanks
Bill

 

[kith]

If you have trouble with the limit concept, you can try the corresponding section in Richard Courant's "What is mathematics?". He doesn't just give a definition and a number of examples but also elaborates quite a bit on why we define it that way and how the notation should be understood.

 

[Potatochip911]

Okay thanks guys I will definitely read these.
Edit: I am noticing something odd when they proved that $\lim_{n\to\infty}\frac{1}{n}=0$. They're making the claim that the sequence $(\frac{1}{n})_{n=1}^{\infty}$ converges although I'm aware this is a divergent sequence from the p-test.

 

[Fredrik]

The pdf and the book that you were recommended look good, but they're saying essentially the same things that I have already told you. They're using the same examples, and of course also a few more. Courant (click this) is using the same explanations that I have already posted in this thread. (You can think of it as a game, and you may find the "all but a finite number of terms" version of the definition much more intuitive). It may help to hear these explanations expressed with a different choice of words, but I don't think that that alone is going to make a difference. I think you just need to look at the definitions, explanations and examples you have already been given, and really think them through.

A goose is said to be a gander if it's an adult male. Suppose that Kevin is a goose. What does the definition of "gander" tell you that the statement "Kevin is a gander" means? The answer is "Kevin is an adult male". I think you had no problems understanding this, and if I change the question by replacing the name "Kevin" with "Michael", I think you would still be able to answer it. But you have so far been unable to do the same with the definition of limit.

It might help to think about what exactly you're doing when you answer the questions about Kevin and Michael. You essentially just substitute the goose from the question into the definition of "gander". And when you've done that once, things get even easier: You can substitute the goose from the next question directly into the answer to the first question, i.e. you just substitute "Michael" for "Kevin" in the sentence "Kevin is an adult male", to get "Michael is an adult male".

The problem with studying books that cover limits of sequences is that none of them is going to explain how to substitute one thing for another in a sentence. That's something that you just need to think about until you get it. I will show you a few examples. I'll start with the most trivial question possible, and then increase the complexity as little as possible in each step.

Definition: Let $(x_n)_{n=1}^\infty$ be a sequence of real numbers. A real number $x$ is said to be a limit of $(x_n)_{n=1}^\infty$, if for all $\varepsilon>0$, there's a positive integer $N$ such that the following implication holds for all positive integers $n$.
$$n\geq N\ \Rightarrow\ |x_n-x|<\varepsilon.$$

Question 1: Let $(x_n)_{n=1}^\infty$ be a sequence of real numbers. Let $x$ be a real number. What does the statement "$x$ is a limit of $(x_n)_{n=1}^\infty$" mean?

Answer:

Spoiler

For all $\varepsilon>0$, there's a positive integer $N$ such that the following implication holds for all positive integers $n$.
$$n\geq N\ \Rightarrow\ |x_n-x|<\varepsilon.$$

Question 2: Let $(x_n)_{n=1}^\infty$ be a sequence of real numbers. Let $y$ be a real number. What does the statement "$y$ is a limit of $(x_n)_{n=1}^\infty$" mean?

Answer:

Spoiler

For all $\varepsilon>0$, there's a positive integer $N$ such that the following implication holds for all positive integers $n$.
$$n\geq N\ \Rightarrow\ |x_n-y|<\varepsilon.$$

Question 3: Let $(y_n)_{n=1}^\infty$ be a sequence of real numbers. Let $x$ be a real number. What does the statement "$x$ is a limit of $(y_n)_{n=1}^\infty$" mean?

Answer:

Spoiler

For all $\varepsilon>0$, there's a positive integer $N$ such that the following implication holds for all positive integers $n$.
$$n\geq N\ \Rightarrow\ |y_n-x|<\varepsilon.$$

Question 4: Let $(y_n)_{n=1}^\infty$ be a sequence of real numbers. Let $y$ be a real number. What does the statement "$y$ is a limit of $(y_n)_{n=1}^\infty$" mean?

Answer:

Spoiler

For all $\varepsilon>0$, there's a positive integer $N$ such that the following implication holds for all positive integers $n$.
$$n\geq N\ \Rightarrow\ |y_n-y|<\varepsilon.$$

Question 5: Let $(x_n)_{n=1}^\infty$ be a sequence of real numbers. Let $c$ and $y$ be real numbers. What does the statement "$cy$ is a limit of $(x_n)_{n=1}^\infty$" mean?

Answer:

Spoiler

For all $\varepsilon>0$, there's a positive integer $N$ such that the following implication holds for all positive integers $n$.
$$n\geq N\ \Rightarrow\ |x_n-cy|<\varepsilon.$$

Question 6: Let $(x_n)_{n=1}^\infty$ be a sequence of real numbers. Let $c$ and $y$ be real numbers. What does the statement "$y$ is a limit of $(cx_n)_{n=1}^\infty$" mean?

Answer:

Spoiler

For all $\varepsilon>0$, there's a positive integer $N$ such that the following implication holds for all positive integers $n$.
$$n\geq N\ \Rightarrow\ |cx_n-y|<\varepsilon.$$

Question 7: Let $(x_n)_{n=1}^\infty$ be a sequence of real numbers. Let $c$ and $x$ be real numbers. What does the statement "$cx$ is a limit of $(cx_n)_{n=1}^\infty$" mean?

Answer:

Spoiler

For all $\varepsilon>0$, there's a positive integer $N$ such that the following implication holds for all positive integers $n$.
$$n\geq N\ \Rightarrow\ |cx_n-cx|<\varepsilon.$$

 

[Potatochip911]

The pdf and the book that you were recommended look good, but they're saying essentially the same things that I have already told you. They're using the same examples, and of course also a few more. Courant (click this) is using the same explanations that I have already posted in this thread. (You can think of it as a game, and you may find the "all but a finite number of terms" version of the definition much more intuitive). It may help to hear these explanations expressed with a different choice of words, but I don't think that that alone is going to make a difference. I think you just need to look at the definitions, explanations and examples you have already been given, and really think them through.

A goose is said to be a gander if it's an adult male. Suppose that Kevin is a goose. What does the definition of "gander" tell you that the statement "Kevin is a gander" means? The answer is "Kevin is an adult male". I think you had no problems understanding this, and if I change the question by replacing the name "Kevin" with "Michael", I think you would still be able to answer it. But you have so far been unable to do the same with the definition of limit.

It might help to think about what exactly you're doing when you answer the questions about Kevin and Michael. You essentially just substitute the goose from the question into the definition of "gander". And when you've done that once, things get even easier: You can substitute the goose from the next question directly into the answer to the first question, i.e. you just substitute "Michael" for "Kevin" in the sentence "Kevin is an adult male", to get "Michael is an adult male".

The problem with studying books that cover limits of sequences is that none of them is going to explain how to substitute one thing for another in a sentence. That's something that you just need to think about until you get it. I will show you a few examples. I'll start with the most trivial question possible, and then increase the complexity as little as possible in each step.

Definition: Let $(x_n)_{n=1}^\infty$ be a sequence of real numbers. A real number $x$ is said to be a limit of $(x_n)_{n=1}^\infty$, if for all $\varepsilon>0$, there's a positive integer $N$ such that the following implication holds for all positive integers $n$.
$$n\geq N\ \Rightarrow\ |x_n-x|<\varepsilon.$$

Question 1: Let $(x_n)_{n=1}^\infty$ be a sequence of real numbers. Let $x$ be a real number. What does the statement "$x$ is a limit of $(x_n)_{n=1}^\infty$" mean?

Answer:

Spoiler

For all $\varepsilon>0$, there's a positive integer $N$ such that the following implication holds for all positive integers $n$.
$$n\geq N\ \Rightarrow\ |x_n-x|<\varepsilon.$$

Question 2: Let $(x_n)_{n=1}^\infty$ be a sequence of real numbers. Let $y$ be a real number. What does the statement "$y$ is a limit of $(x_n)_{n=1}^\infty$" mean?

Answer:

Spoiler

For all $\varepsilon>0$, there's a positive integer $N$ such that the following implication holds for all positive integers $n$.
$$n\geq N\ \Rightarrow\ |x_n-y|<\varepsilon.$$

Question 3: Let $(y_n)_{n=1}^\infty$ be a sequence of real numbers. Let $x$ be a real number. What does the statement "$x$ is a limit of $(y_n)_{n=1}^\infty$" mean?

Answer:

Spoiler

For all $\varepsilon>0$, there's a positive integer $N$ such that the following implication holds for all positive integers $n$.
$$n\geq N\ \Rightarrow\ |y_n-x|<\varepsilon.$$

Question 4: Let $(y_n)_{n=1}^\infty$ be a sequence of real numbers. Let $y$ be a real number. What does the statement "$y$ is a limit of $(y_n)_{n=1}^\infty$" mean?

Answer:

Spoiler

For all $\varepsilon>0$, there's a positive integer $N$ such that the following implication holds for all positive integers $n$.
$$n\geq N\ \Rightarrow\ |y_n-y|<\varepsilon.$$

Question 5: Let $(x_n)_{n=1}^\infty$ be a sequence of real numbers. Let $c$ and $y$ be real numbers. What does the statement "$cy$ is a limit of $(x_n)_{n=1}^\infty$" mean?

Answer:

Spoiler

For all $\varepsilon>0$, there's a positive integer $N$ such that the following implication holds for all positive integers $n$.
$$n\geq N\ \Rightarrow\ |x_n-cy|<\varepsilon.$$

Question 6: Let $(x_n)_{n=1}^\infty$ be a sequence of real numbers. Let $c$ and $y$ be real numbers. What does the statement "$y$ is a limit of $(cx_n)_{n=1}^\infty$" mean?

Answer:

Spoiler

For all $\varepsilon>0$, there's a positive integer $N$ such that the following implication holds for all positive integers $n$.
$$n\geq N\ \Rightarrow\ |cx_n-y|<\varepsilon.$$

Question 7: Let $(x_n)_{n=1}^\infty$ be a sequence of real numbers. Let $c$ and $x$ be real numbers. What does the statement "$cx$ is a limit of $(cx_n)_{n=1}^\infty$" mean?

Answer:

Spoiler

For all $\varepsilon>0$, there's a positive integer $N$ such that the following implication holds for all positive integers $n$.
$$n\geq N\ \Rightarrow\ |cx_n-cx|<\varepsilon.$$

So the part where I kept saying "a gander is a gander" is when I was using the limit inside the definition of a limit? I a bit confused as to why we are defining the limits as natural numbers now when before we would define the limit to be $\lim_{n}S_{n}$, for example in post #46 $$|S_{n}-S|<\varepsilon$$ where $\lim_{n}S_{n}=S$

 

[Fredrik]

So the part where I kept saying "a gander is a gander" is when I was using the limit inside the definition of a limit?

Yes. When I asked you to use the definition of limit to explain what it means to say that a real number x is a limit of some specific sequence S, you kept using notations that refer to the limit of S. You certainly can't refer to the limit of S in the statement that's supposed to explain what a limit of S is, any more than you can use the term "gander" in the explanation of what "Kevin is a gander" means.

I a bit confused as to why we are defining the limits as natural numbers now when before we would define the limit to be $\lim_{n}S_{n}$, for example in post #46 $$|S_{n}-S|<\varepsilon$$ where $\lim_{n}S_{n}=S$

Real numbers, not natural numbers. And nothing has changed. The limit of a sequence of real numbers has always been a real number. What else would it be? $\lim_n S_n$ is just a notation for the real number that happens to be the limit of $(S_n)_{n=1}^\infty$. And that notation certainly can't be a part of the statement that defines what a limit of a sequence is. That would be absurd for the reason I just mentioned, and also because it makes no sense to refer to "the" limit of a sequence until you have proved that there's at most one. (This is something you can't do that until after you have defined what a limit of a sequence is).

I didn't use a phrase like "where $\lim_n S_n=S$" in post #46. That wouldn't have made sense in this context. The statement "$\lim_n S_n=S$" means that the real number S is the limit of $(S_n)_{n=1}^\infty$, and post #46 explains what it means to say that S is a limit of $(S_n)_{n=1}^\infty$.

Do you understand the questions and answers in my previous post? Were you able to answer them before you clicked the spoiler buttons?

 

[Potatochip911]

Yes. When I asked you to use the definition of limit to explain what it means to say that a real number x is a limit of some specific sequence S, you kept using notations that refer to the limit of S. You certainly can't refer to the limit of S in the statement that's supposed to explain what a limit of S is, any more than you can use the term "gander" in the explanation of what "Kevin is a gander" means.Real numbers, not natural numbers. And nothing has changed. The limit of a sequence of real numbers has always been a real number. What else would it be? $\lim_n S_n$ is just a notation for the real number that happens to be the limit of $(S_n)_{n=1}^\infty$. And that notation certainly can't be a part of the statement that defines what a limit of a sequence is. That would be absurd for the reason I just mentioned, and also because it makes no sense to refer to "the" limit of a sequence until you have proved that there's at most one. (This is something you can't do that until after you have defined what a limit of a sequence is).

I didn't use a phrase like "where $\lim_n S_n=S$" in post #46. That wouldn't have made sense in this context. The statement "$\lim_n S_n=S$" means that the real number S is the limit of $(S_n)_{n=1}^\infty$, and post #46 explains what it means to say that S is a limit of $(S_n)_{n=1}^\infty$.

Do you understand the questions and answers in my previous post? Were you able to answer them before you clicked the spoiler buttons?

Yea they're all the same except for the variables so I was able to answer them all.

 

[Fredrik]

Yea they're all the same except for the variables so I was able to answer them all.

That's great. Are you now able to complete the proof of the claim that if c and x are real numbers and x is a limit of $(x_n)_{n=1}^\infty$, then cx is a limit of $(cx_n)_{n=1}^\infty$? (I'm avoiding the "lim" notation this time, because it seems to have caused some confusion).

The proof is only two sentences long. The second sentence is very similar to the "for all ε" statement that by the definition of "limit" is equivalent to "cx is a limit of $(cx_n)_{n=1}^\infty$". That sentence involves the variable N. The first sentence uses the assumption that x is a limit of $(x_n)_{n=1}^\infty$ to assign an appropriate value to that N.

 

[Potatochip911]

That's great. Are you now able to complete the proof of the claim that if c and x are real numbers and x is a limit of $(x_n)_{n=1}^\infty$, then cx is a limit of $(cx_n)_{n=1}^\infty$? (I'm avoiding the "lim" notation this time, because it seems to have caused some confusion).

The proof is only two sentences long. The second sentence is very similar to the "for all ε" statement that by the definition of "limit" is equivalent to "cx is a limit of $(cx_n)_{n=1}^\infty$". That sentence involves the variable N. The first sentence uses the assumption that x is a limit of $(x_n)_{n=1}^\infty$ to assign an appropriate value to that N.

Let $\varepsilon>0$, let $c \in\mathbb{R}$, let x be a limit of the sequence $(x_{n})_{n=1}^{\infty}$ and let $N$ be a positive integer such that the following implication holds for all positive integers $n$
$$
n\geq N \Rightarrow |x_{n}-x|<\frac{\varepsilon}{|c|} \\
|c||x_{n}-x|=|cx_{n}-cx|<|c|\frac{\varepsilon}{|c|}
$$

 

[Fredrik]

Let $\varepsilon>0$, let $c \in\mathbb{R}$, let x be a limit of the sequence $(x_{n})_{n=1}^{\infty}$ and let $N$ be a positive integer such that the following implication holds for all positive integers $n$
$$
n\geq N \Rightarrow |x_{n}-x|<\frac{\varepsilon}{|c|}$$

This part is perfect.

$$|c||x_{n}-x|=|cx_{n}-cx|<|c|\frac{\varepsilon}{|c|}$$

You need to make this calculation part of a statement that (together with the fact that ε is an arbitrary positive real number) implies that cx is a limit of $(cx_n)_{n=1}^\infty$.

Also, the calculation is easier to read if you write
$$|cx_{n}-cx| =|c||x_{n}-x|<|c|\frac{\varepsilon}{|c|}$$ instead.