Prove Sum Approximation Theorem

[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.

 

[Potatochip911]

This part is perfect.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.

Hmm, I'm not entirely sure what to say. Does it have to do with all but a finite number of terms are within epsilon?

 

[Fredrik]

Hmm, I'm not entirely sure what to say. Does it have to do with all but a finite number of terms are within epsilon?

Yes it does. But keep in mind that the standard definition is what defines the precise meaning of "all but a finite number of terms". So you should use the type of language that's used in the standard definition and avoid the phrase "all but a finite".

You also need to keep in mind that a sentence (like $|cx_n-cx|=|c||x_n-x|$) isn't really a statement unless every variable in it has been assigned a value or is the target of a "for all" or a "there exists". The words "let $c\in\mathbb R$, let x be a limit of..." take care of x and c, but what about n?

The reason the "let" statements "take care of" x and c is that saying "Let $x\in\mathbb R$. (Some claim about x)" is equivalent to saying "For all $x\in\mathbb R$, we have (that same claim about x)". The "let" is just a trick to allow us to break a "for all" statement up into multiple sentences.

The statement that finishes the proof is something that is guaranteed to be true by our choice of N, and that would have been false if $cx$ hadn't been a limit of $(cx_n)_{n=1}^\infty$. That last part is the reason why it's so essential to know what exactly it means to say that $cx$ is a limit of $(cx_n)_{n=1}^\infty$.

 

[Potatochip911]

Yes it does. But keep in mind that the standard definition is what defines the precise meaning of "all but a finite number of terms". So you should use the type of language that's used in the standard definition and avoid the phrase "all but a finite".

You also need to keep in mind that a sentence (like $|cx_n-cx|=|c||x_n-x|$) isn't really a statement unless every variable in it has been assigned a value or is the target of a "for all" or a "there exists". The words "let $c\in\mathbb R$, let x be a limit of..." take care of x and c, but what about n?

The reason the "let" statements "take care of" x and c is that saying "Let $x\in\mathbb R$. (Some claim about x)" is equivalent to saying "For all $x\in\mathbb R$, we have (that same claim about x)". The "let" is just a trick to allow us to break a "for all" statement up into multiple sentences.

The statement that finishes the proof is something that is guaranteed to be true by our choice of N, and that would have been false if $cx$ hadn't been a limit of $(cx_n)_{n=1}^\infty$. That last part is the reason why it's so essential to know what exactly it means to say that $cx$ is a limit of $(cx_n)_{n=1}^\infty$.

Let $n$ be a positive integer such that for $n\geq N$ the following holds,
$$|cx_{n}-cx|=|c||x_{n}-x|<|c|\frac{\varepsilon}{|c|}$$
Is this correct?

 

[Fredrik]

Let $n$ be a positive integer such that for $n\geq N$ the following holds,
$$|cx_{n}-cx|=|c||x_{n}-x|<|c|\frac{\varepsilon}{|c|}$$
Is this correct?

Not quite. You made it sound like you're assuming that the calculation is valid instead of showing that it is.

You should also add a step to the calculation that simplifies the final result.

 

[Potatochip911]

Not quite. You made it sound like you're assuming that the calculation is valid instead of showing that it is.

You should also add a step to the calculation that simplifies the final result.

It sounds like assuming the calculation is valid because I left out post #89 or am I missing something? As for something that simplifies the final result I will have to think about that some more.

 

[Fredrik]

It sounds like an assumption because you let n be an integer such that the equalities and the inequality in the next line hold. Can you let n be an integer that satisfies some other requirement, and then show that those equalities and that inequality hold for such an n?

Can you simplify $2\frac x 2$?

 

[Potatochip911]

It sounds like an assumption because you let n be an integer such that the equalities and the inequality in the next line hold. Can you let n be an integer that satisfies some other requirement, and then show that those equalities and that inequality hold for such an n?

Can you simplify $2\frac x 2$?

I'm not very sure of what I'm doing but does it make sense to work backwards from $|cx_{n}-cx|<\varepsilon$ to $|x_{n}-x|<\frac{\varepsilon}{|c|}$ which is something that we know is valid? I'm concerned this is similar to the mistakes I made in post #57? And I see now that the simplification is $|c|\frac{\varepsilon}{|c|}=\varepsilon$

 

[Fredrik]

I'm not very sure of what I'm doing but does it make sense to work backwards from $|cx_{n}-cx|<\varepsilon$ to $|x_{n}-x|<\frac{\varepsilon}{|c|}$ which is something that we know is valid? I'm concerned this is similar to the mistakes I made in post #57? And I see now that the simplification is $|c|\frac{\varepsilon}{|c|}=\varepsilon$

Yes, that's the simplification I had in mind. I don't think you're making any of the mistakes you made in post #57. I don't know how to explain this in a way that enables you to find the right way to finish the proof for yourself, so I'll just show you the proof now.

Let ε>0. Let N be a positive integer such that the following implication holds for all positive integers n.
$$n\geq N\ \Rightarrow\ |x_n-x|<\frac{\varepsilon}{|c|}.$$ Let n be a positive integer such that $n\geq N$. We have
$$|cx_n-cx|=|c||x_n-x|<|c|\frac{\varepsilon}{|c|}=\varepsilon.$$

That's the entire proof. It proves the following claim:

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

Note that the proof starts by letting ε be an arbitrary positive real number. It has to, because the statement we want to prove is of the type "for all positive real numbers ε..." Then we assign a value to N. Note that this specific assignment is possible because we have assumed that x is a limit of $(x_n)_{n=1}^\infty$. Now we just need to prove that the implication in the statement that we want to prove holds for all positive integers n. So we either let n be an arbitrary positive integer and prove the implication, or (equivalently) let n be an arbitrary positive integer such that $n\geq N$ and prove the inequality.

 

[Potatochip911]

Yes, that's the simplification I had in mind. I don't think you're making any of the mistakes you made in post #57. I don't know how to explain this in a way that enables you to find the right way to finish the proof for yourself, so I'll just show you the proof now.

Let ε>0. Let N be a positive integer such that the following implication holds for all positive integers n.
$$n\geq N\ \Rightarrow\ |x_n-x|<\frac{\varepsilon}{|c|}.$$ Let n be a positive integer such that $n\geq N$. We have
$$|cx_n-cx|=|c||x_n-x|<|c|\frac{\varepsilon}{|c|}=\varepsilon.$$

That's the entire proof. It proves the following claim:

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

Note that the proof starts by letting ε be an arbitrary positive real number. It has to, because the statement we want to prove is of the type "for all positive real numbers ε..." Then we assign a value to N. Note that this specific assignment is possible because we have assumed that x is a limit of $(x_n)_{n=1}^\infty$. Now we just need to prove that the implication in the statement that we want to prove holds for all positive integers n. So we either let n be an arbitrary positive integer and prove the implication, or (equivalently) let n be an arbitrary positive integer such that $n\geq N$ and prove the inequality.

Okay I think I'm starting to understand this.

 

[Fredrik]

Okay I think I'm starting to understand this.

That's good news. I will not pressure you to do more proofs here, but I strongly recommend that you take this opportunity to do some exercises on this topic from any book that covers it. If you get stuck on a problem, I suggest that you start a new thread about it.

I will show you my proof of the theorem you asked about in post #1, as well as the proof of the result that no sequence has more than one limit. We will need the following lemma.

Lemma: Let x and y be real numbers. If x<y+ε for all ε>0, then x≤y.

Proof: Suppose that x<y+ε for all ε>0. We will prove that x≤y by deriving a falsehood from the assumption that this is false. So suppose that y<x. We have (x-y)/2>0 and therefore
$$x<y+\frac{x-y}{2} =\frac{x+y}{2} <\frac{x+x}{2} =x.$$ The inequality x<x is clearly a falsehood.

Theorem: If x and y are both limits of $(x_n)_{n=1}^\infty$, then $x=y$.

Proof:
Let ε>0. Let $N_1$ be a positive integer such that the following implication holds for all positive integers n.
$$n\geq N_1\ \Rightarrow\ |x_n-x|<\frac{\varepsilon}{2}.$$ Let $N_2$ be a positive integer such that the following implication holds for all positive integers n.
$$n\geq N_2\ \Rightarrow\ |x_n-y|<\frac{\varepsilon}{2}.$$ Define $n=\max\{N_1,N_2\}$. We have
$$|x-y|=|x-x_n+x_n-y|\leq|x-x_n|+|x_n-y|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon.$$ Since ε is an arbitrary positive real number, this implies that |x-y|≤0. (See the lemma). Since |x-y|≥0, this implies that |x-y|=0. This implies that x-y=0. This implies that x=y.

Comment: This result is the reason why it makes sense to talk about "the" limit of a convergent sequence instead of "a" limit.

Theorem: Let x be a real number such that |x|<1. Let $(a_n)_{n=0}^\infty$ be a sequence of real numbers such that $|a_n|<|a_{n+1}|$ for all non-negative integers n, and $\sum_{n=0}^\infty a_n x^n$ is convergent. Define $S=\sum_{n=0}^\infty a_n x^n$. For all non-negative integers N, we have
$$\left|S-\sum_{n=0}^N a_n x^n\right|<\frac{\left|a_{N+1} x^{N+1}\right|}{1-|x|}.$$

Proof: Let ε>0. Let N be a non-negative integer. Let M be a positive integer such that the following implication holds for all positive integers m,
$$m\geq M\ \Rightarrow\ \left|\sum_{n=0}^m a_n x^n-S\right|<\varepsilon.$$ Let m be an integer greater than max{M,N}. We have
\begin{align*}
&\left|S-\sum_{n=0}^N a_n x^n\right| =\left|S-\sum_{n=0}^m a_n x_n +\sum_{n=N+1}^m a_n x^n\right|
\leq \left|S-\sum_{n=0}^m a_n x_n\right| + \left|\sum_{n=N+1}^m a_n x^n\right|\\
&<\varepsilon+\left|\sum_{n=N+1}^m a_n x^n\right| \leq \varepsilon +\sum_{n=N+1}^m |a_n x^n| = \varepsilon + \left|a_{N+1} x^{N+1}\right| \left(1+\frac{|a_{N+2}|}{|a_{N+1}|}|x|+\cdots+\frac{|a_m|}{|a_{N+1}|} |x|^{m-N-1}\right)\\
&<\varepsilon +\left|a_{N+1} x^{N+1}\right|\left(1+|x|+\cdots+|x|^{m-N-1}\right) =\varepsilon +\left|a_{N+1} x^{N+1}\right| \frac{\left(1+|x|+\cdots+|x|^{m-N-1}\right)(1-|x|)}{1-|x|}\\
&=\varepsilon+\left|a_{N+1}x^{N+1}\right| \frac{1-|x|^{m-N}}{1-|x|} <\varepsilon+\left|a_{N+1}x^{N+1}\right| \frac{1}{1-|x|}.
\end{align*}
Since $\varepsilon$ is an arbitrary positive real number and N is an arbitrary non-negative integer, this implies that
$$\left|S-\sum_{n=0}^N a_n x^n\right| \leq \frac{\left|a_{N+1}x^{N+1}\right|}{1-|x|}$$ for all non-negative integers N.