Prove Sum Approximation Theorem

[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.