Prove Sum Approximation Theorem

[Potatochip911]

Homework Statement

I put up the image so that you can see the hints if you're curious. I am supposed to prove that if $S=\sum_{n=0}^{\infty}a_{n}x^{n}$ converges for $|x|<1$, and if $|a_{n+1}|<|a_{n}|$ for $n>N$, then $$|S-\sum_{n=0}^{N}a_{n}x^{n}|<|a_{N+1}x^{N+1}|\div (1-|x|)$$

Homework Equations

3. The Attempt at a Solution [/B]
$$S=\sum_{n=0}^{N}a_{n}x^{n}+\sum_{N+1}^{\infty}a_{n}x^{n} \\
S\leq \sum_{n=0}^{N}a_{n}x^{n}+|a_{N+1}x^{N+1}|+|a_{N+2}x^{N+2}|+\cdots \\
S<\sum_{n=0}^{N}a_{n}x^{n}+|a_{N+1}x^{N+1}|+|a_{N+1}x^{N+2}+\cdots \\
S<\sum_{n=0}^{N}a_{n}x^{n}+(|a_{N+1}|)(|x^{N+1}|+|x^{N+2}|+\cdots) \\
S-\sum_{n=0}^{N}a_{n}x^{n}<(|a_{N+1}|)(\frac{1}{1-|x|})
$$
So as you can see I haven't quite gotten it, the main problem is that I'm missing the $x^{N+1}$ on the RHS.

 

[Fredrik]

Think about what the statement $\sum_{n=0}^\infty a_n x^n=S$ means. I think the first thing you do should be to use that to rewrite $|S-\sum_{n=0}^N a_n x^n|$.

 

[Potatochip911]

Think about what the statement $\sum_{n=0}^\infty a_n x^n=S$ means. I think the first thing you do should be to use that to rewrite $|S-\sum_{n=0}^N a_n x^n|$.

Okay so that fixed the problems with the absolute values but I'm still not getting that $x^{N+1}$ term.
$$S=\sum_{n=0}^{\infty}a_{n}x^{n} \\
S=\sum_{n=0}^{N}a_{n}x^{n}+\sum_{n=N+1}^{\infty} a_{n}x^{n}\\
|S-\sum_{n=0}^{N}a_{n}x^{n}|=|\sum_{n=N+1}^{\infty}a_{n}x^{n}| \\
|S-\sum_{n=0}^{N}a_{n}x^{n}| \leq |a_{N+1}x^{N+1}|+|a_{N+2}x^{N+2}|+\cdots \\
|S-\sum_{n=0}^{N}a_{n}x^{n}| < |a_{N+1}x^{N+1}|+|a_{N+1}x^{N+2}|+\cdots \\
|S-\sum_{n=0}^{N}a_{n}x^{n}| < (|a_{N+1}|)(|x^{N+1}|+|x^{N+2}|+\cdots) \\
|S-\sum_{n=0}^{N}a_{n}x^{n}| < (|a_{N+1})(\lim_{n\to\infty} \frac{1-|x|^{n}}{1-|x|})
$$
and since $|x|<1$ as $n\to\infty, x\to 0$ so
$$|S-\sum_{n=0}^{N}a_{n}x^{n}| < (|a_{N+1})(\frac{1}{1-|x|})$$

 

[Dick]

Okay so that fixed the problems with the absolute values but I'm still not getting that $x^{N+1}$ term.
$$S=\sum_{n=0}^{\infty}a_{n}x^{n} \\
S=\sum_{n=0}^{N}a_{n}x^{n}+\sum_{n=N+1}^{\infty} a_{n}x^{n}\\
|S-\sum_{n=0}^{N}a_{n}x^{n}|=|\sum_{n=N+1}^{\infty}a_{n}x^{n}| \\
|S-\sum_{n=0}^{N}a_{n}x^{n}| \leq |a_{N+1}x^{N+1}|+|a_{N+2}x^{N+2}|+\cdots \\
|S-\sum_{n=0}^{N}a_{n}x^{n}| < |a_{N+1}x^{N+1}|+|a_{N+1}x^{N+2}|+\cdots \\
|S-\sum_{n=0}^{N}a_{n}x^{n}| < (|a_{N+1}|)(|x^{N+1}|+|x^{N+2}|+\cdots) \\
|S-\sum_{n=0}^{N}a_{n}x^{n}| < (|a_{N+1})(\lim_{n\to\infty} \frac{1-|x|^{n}}{1-|x|})
$$
and since $|x|<1$ as $n\to\infty, x\to 0$ so
$$|S-\sum_{n=0}^{N}a_{n}x^{n}| < (|a_{N+1})(\frac{1}{1-|x|})$$

$\frac{1}{1-|x|}$ is the sum of $(1+|x|+|x^2|+\cdots)$. You have $(|x^{N+1}|+|x^{N+2}|+\cdots)$. Does that tell you where the missing factor is?

 

[Fredrik]

Okay so that fixed the problems with the absolute values but I'm still not getting that $x^{N+1}$ term.
$$S=\sum_{n=0}^{N}a_{n}x^{n}+\sum_{n=N+1}^{\infty} a_{n}x^{n}$$

I'm not super comfortable with this rewrite. I'm not saying that it's wrong, only that it requires a separate epsilon-delta proof. Also, the manipulations you're doing later forces the reader to think at each step about whether the step is valid, since you're dealing with the sum of an infinite sequence. The proof would be much nicer if you could find a way to work with a sum of a finite sequence. I will elaborate on that below.

$$|S-\sum_{n=0}^{N}a_{n}x^{n}|=|\sum_{n=N+1}^{\infty}a_{n}x^{n}| \\
|S-\sum_{n=0}^{N}a_{n}x^{n}| \leq |a_{N+1}x^{N+1}|+|a_{N+2}x^{N+2}|+\cdots \\
|S-\sum_{n=0}^{N}a_{n}x^{n}| < |a_{N+1}x^{N+1}|+|a_{N+1}x^{N+2}|+\cdots \\
|S-\sum_{n=0}^{N}a_{n}x^{n}| < (|a_{N+1}|)(|x^{N+1}|+|x^{N+2}|+\cdots)
$$

Why not move $|x^{N+1}|$ outside as well? Actually, you have to do that if you are going to use a formula for a geometric series with 1 as the first term..

I guess I should elaborate a bit on what I had in mind when I gave you that hint in my previous post. The assumption that $\sum_{n=0}^\infty a_n x^n=S$ means that for all $\varepsilon>0$ there's a positive integer $M$ such that the following implication holds for all positive integers m:
$$m\geq M\ \Rightarrow\ \left|\sum_{n=0}^m a_n x^n-S\right|<\varepsilon.$$ This is the result I wanted you to use to prove a result of the form $\big|S-\sum_{n=0}^N a_nx^n\big|<\text{something}$. The right-hand side will involve a sum with a finite number of terms, and if you rewrite it in the form $A(1+\dots)$, then everything works out nicely.

 

[Potatochip911]

$\frac{1}{1-|x|}$ is the sum of $(1+|x|+|x^2|+\cdots)$. You have $(|x^{N+1}|+|x^{N+2}|+\cdots)$. Does that tell you where the missing factor is?

Haha yea that was quite a dumb mistake I made there.

I'm not super comfortable with this rewrite. I'm not saying that it's wrong, only that it requires a separate epsilon-delta proof. Also, the manipulations you're doing later forces the reader to think at each step about whether the step is valid, since you're dealing with the sum of an infinite sequence. The proof would be much nicer if you could find a way to work with a sum of a finite sequence. I will elaborate on that below.Why not move $|x^{N+1}|$ outside as well? Actually, you have to do that if you are going to use a formula for a geometric series with 1 as the first term..

I guess I should elaborate a bit on what I had in mind when I gave you that hint in my previous post. The assumption that $\sum_{n=0}^\infty a_n x^n=S$ means that for all $\varepsilon>0$ there's a positive integer $M$ such that the following implication holds for all positive integers m:
$$m\geq M\ \Rightarrow\ \left|\sum_{n=0}^m a_n x^n-S\right|<\varepsilon.$$ In particular, if $M$ is such an integer, we have
$$\left|\sum_{n=0}^M a_n x^n-S\right|<\varepsilon.$$ This is the result I wanted you to use to find a number greater than $\big|S-\sum_{n=0}^N a_nx^n\big|$. (Suppose that $M>N$). The result will involve a sum with a finite number of terms, and if you rewrite it in the form $A(1+\dots)$, then everything works out nicely.

I realize you have already given me a lot of hints but I'm unfortunately failing to find a way to get a finite sequence.

 

[Fredrik]

I realize you have already given me a lot of hints but I'm unfortunately failing to find a way to get a finite sequence.

Not sure what I can tell you other than that it's not very complicated. I found this proof just by thinking about in the bathroom right after I had read your original post. Didn't even need to use a pen.

I guess I can tell you that it's a pretty straightforward way to use the definition of the statement $\sum_{n=0}^\infty a_n x^n =S$ to obtain a result of the form $\big|S-\sum_{n=0}^N a_nx^n\big|<\text{something}$, and that (like almost all epsilon-delta proofs) it involves the triangle inequality.

 

[Potatochip911]

Not sure what I can tell you other than that it's not very complicated. I found this proof just by thinking about in the bathroom right after I had read your original post. Didn't even need to use a pen.

I guess I can tell you that it's a pretty straightforward way to use the definition of the statement $\sum_{n=0}^\infty a_n x^n =S$ to obtain a result of the form $\big|S-\sum_{n=0}^N a_nx^n\big|<\text{something}$, and that (like almost all epsilon-delta proofs) it involves the triangle inequality.

And it doesn't involve $S=\sum_{n=0}^{N}a_{n}x^{n}+\sum_{N+1}^{\infty}a_{n}x^{n}$ right?

 

[Fredrik]

And it doesn't involve $S=\sum_{n=0}^{N}a_{n}x^{n}+\sum_{N+1}^{\infty}a_{n}x^{n}$ right?

Right. The key step is to use the definition of the statement $\sum_{n=0}^\infty a_n x^n=S$.

 

[ChrisVer]

Can I ask something which I don't find clear why you are doing it?
Why do you write $a_{N+2}= a_{N+1}$?

*edit PS: I got why you did that. Forget the question*

 

[Potatochip911]

Right. The key step is to use the definition of the statement $\sum_{n=0}^\infty a_n x^n=S$.

So $$a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots+a_{m}x^{m}=S; \space \space \space \space m\to\infty $$
I looked over this link Triangle Inequality Theorem, I just don't understand how to use it to get an $a_{N+1}$ term when I am not separating the sum into two separate sums. I tried something like $$a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots+a_{N}x^{N}<S \space \space \space \space m>N$$ but this result seems useless and also seems to render the triangle inequality useless since I already have an inequality now.

 

[Fredrik]

So $$a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots+a_{m}x^{m}=S; \space \space \space \space m\to\infty $$

Yes, the expression $\lim_{m\to\infty}\sum_{n=0}^m a_n x^n=S$ can be viewed as the definition of what $\sum_{n=0}^\infty a_n x^n=S$ means, but it can also be viewed as a trivial change of notation. What I wanted you to use is the definition of $\lim_{m\to\infty}\sum_{n=0}^m a_n x^n=S$. What does that statement mean? (I posted the answer to that in post #5).

...when I am not separating the sum into two separate sums.

I only said that you shouldn't rewrite the infinite series $\sum_{n=0}^\infty a_n x^n$ as finite series plus an infinite series. That infinite series isn't even present in the expression that you need to evalutate: $\big|S-\sum_{n=0}^N a_n x^n\big|$. Yes, there's an S in there, and it represents the same number as $\sum_{n=0}^\infty a_n x^n$, but replacing the S with $\sum_{n=0}^\infty a_n x^n$ will only take you further from the solution.

I tried something like...

I don't see an $\varepsilon$ in any of your attempts. My suggestion was to use a specific definition, which involves an arbitrary positive real number that's traditionally denoted by $\varepsilon$. If an $\varepsilon$ hasn't shown up in your calculations, you still haven't used that definition.

In this context, the triangle inequality is just the following statement: For all real numbers x and y, we have $|x+y|\leq|x|+|y|$. Edit: You also need its (rather trivial) generalization to sums with a finite number of terms: For all positive integers n, and all sequences $(x_i)_{i=1}^n$ in $\mathbb R$, we have $\big|\sum_{i=1}^n x_i\big|\leq\sum_{i=1}^n|x_i|.$

 

[ChrisVer]

The right-hand side will involve a sum with a finite number of terms

Are you sure that the sum in the RHS will involve a finite number of terms? In particular that $\frac{1}{1+|x|}$ really calls in my mind that you have to use the geometric series.

 

[Fredrik]

Are you sure that the sum in the RHS will involve a finite number of terms? In particular that $\frac{1}{1+|x|}$ really calls in my mind that you have to use the geometric series.

A sum with a finite number of terms will show up almost immediately. The argument for why it's small enough may however involve a full geometric series.

I will PM you the details, since I can't post a full solution here.

 

[momoko]

Your solution in #1 is correct until the last line. Just use the correct formula for GS (=a/(1-r)), then you can complete the proof.

 

[Fredrik]

Yes, I should have stated more clearly that the approach in post #1 isn't wrong (until the last step, where the formula for the sum of a geometric series was applied incorrectly). I still like my method better because the calculations in posts #1 and #3 use a couple of theorems without saying anything about it:

Theorem 1:

If $\sum_{n=0}^\infty b_n$ is convergent, then for all positive integers N, so is $\sum_{n=N+1}^\infty b_n$, and we have
$$\sum_{n=0}^\infty b_n=\sum_{n=0}^N b_n +\sum_{n=N+1}^\infty b_n.$$
Theorem 2:

If $\sum_{n=0}^\infty b_n$ is convergent and k is a real number, then $\sum_{n=0}^\infty kb_n$ is convergent, and we have
$$\sum_{n=0}^\infty kb_n =k\sum_{n=0}^\infty b_n.$$
These theorems can be considered elementary enough that they don't need to be proved or even referenced in a proof like this, but I'm getting the impression that Potatochip911 doesn't yet have the experience with proofs that involve the definition of "limit" to find these theorems and their proofs simple.

 

[Potatochip911]

Yes, I should have stated more clearly that the approach in post #1 isn't wrong (until the last step, where the formula for the sum of a geometric series was applied incorrectly). I still like my method better because the calculations in posts #1 and #3 use a couple of theorems without saying anything about it:

Theorem 1:

If $\sum_{n=0}^\infty b_n$ is convergent, then for all positive integers N, so is $\sum_{n=N+1}^\infty b_n$, and we have
$$\sum_{n=0}^\infty b_n=\sum_{n=0}^N b_n +\sum_{n=N+1}^\infty b_n.$$
Theorem 2:

If $\sum_{n=0}^\infty b_n$ is convergent and k is a real number, then $\sum_{n=0}^\infty kb_n$ is convergent, and we have
$$\sum_{n=0}^\infty kb_n =k\sum_{n=0}^\infty b_n.$$
These theorems can be considered elementary enough that they don't need to be proved or even referenced in a proof like this, but I'm getting the impression that Potatochip911 doesn't yet have the experience with proofs that involve the definition of "limit" to find these theorems and their proofs simple.

Yea sorry for not replying in a long time but I don't have any experience with epsilon-delta proofs for limits so I went looking for that in my calculus textbook although I haven't finished the problems on them yet.

 

[Fredrik]

I don't have any experience with epsilon-delta proofs for limits

I suspected that, since you found my hint hard to follow and none of your posts contained the symbol $\varepsilon$.

Don't worry about limits of functions right now. Focus on the definition of limits of sequences. It looks like this:

Definition: Let $(x_n)_{n=1}^\infty$ be a sequence of real numbers. A real number L 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.
$$n\geq N\ \Rightarrow\ |x_n-L|<\varepsilon.$$
This is a slightly less formal way to say the same thing: L is a limit of a sequence, if for all $\varepsilon>0$, all but a finite number of terms of the sequence are at a distance from L that's less than $\varepsilon$. (If you're concerned about what "all but a finite number" means, just look at the first version of the definition again. The difference between these two versions of the definition is that the first one spells out exactly what "all but a finite number" means, while the second one just uses that phrase without explanation).

Some people find the following description useful. Imagine that we're playing the following game. I get to pick the value of $\varepsilon$. It has to be a positive real number, but there are no other restrictions. I have to tell you what value I picked. Then you get to pick the value of N. It has to be a positive integer, but there are no other restrictions. You win the game if all the $x_n$ with $n\geq N$ are at a distance from L that's less than $\varepsilon$. What the definition is saying is the following statements are equivalent:

1. L is a limit of this sequence.
2. No matter what value I assigned to $\varepsilon$, you will win the game (assuming that you play perfectly).

Are you able to use the definition to prove that $\lim_{n\to\infty}\frac 1 n=0$? Give it a try before you look at my solution below:

Spoiler

The proof looks like this:

Let $\varepsilon>0$. Let $N$ be a positive integer such that $N>1/\varepsilon$. Let $n$ be a positive integer such that $n\geq N$. We have
$$\left|\frac 1 n-0\right|=\frac 1 n\leq \frac 1 N < \varepsilon.$$

That's the complete proof. Finding the proof is another matter. I will explain my process. Since the statement you want to prove is of the form "for all positive real numbers $\varepsilon$...", the proof should start with some version of the statement "Let $\varepsilon$ be an arbitrary positive real number". I think it's OK to simplify it to "Let $\varepsilon>0$".

So the first sentence of the proof writes itself. To continue, you need to figure out how to define your N. You want the inequality $\big|\frac 1 n -0\big|<\varepsilon$ to hold for all n such that $n\geq N$. That means in particular that you want the inequality to hold when n=N. So we need an N such that $\frac 1 N <\varepsilon$. This inequality is equivalent to $N>1/\varepsilon$. That explains my choice of N. The rest is straightforward.

 

[Potatochip911]

I suspected that, since you found my hint hard to follow and none of your posts contained the symbol $\varepsilon$.

Don't worry about limits of functions right now. Focus on the definition of limits of sequences. It looks like this:

Definition: Let $(x_n)_{n=1}^\infty$ be a sequence of real numbers. A real number L 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.
$$n\geq N\ \Rightarrow\ |x_n-L|<\varepsilon.$$
This is a slightly less formal way to say the same thing: L is a limit of a sequence, if for all $\varepsilon>0$, all but a finite number of terms of the sequence are at a distance from L that's less than $\varepsilon$. (If you're concerned about what "all but a finite number" means, just look at the first version of the definition again. The difference between these two versions of the definition is that the first one spells out exactly what "all but a finite number" means, while the second one just uses that phrase without explanation).

Some people find the following description useful. Imagine that we're playing the following game. I get to pick the value of $\varepsilon$. It has to be a positive real number, but there are no other restrictions. I have to tell you what value I picked. Then you get to pick the value of N. It has to be a positive integer, but there are no other restrictions. You win the game if all the $x_n$ with $n\geq N$ are at a distance from L that's less than $\varepsilon$. What the definition is saying is the following statements are equivalent:

1. L is a limit of this sequence.
2. No matter what value I assigned to $\varepsilon$, you will win the game (assuming that you play perfectly).

Are you able to use the definition to prove that $\lim_{n\to\infty}\frac 1 n=0$? Give it a try before you look at my solution below:

Spoiler

The proof looks like this:

Let $\varepsilon>0$. Let $N$ be a positive integer such that $N>1/\varepsilon$. Let $n$ be a positive integer such that $n\geq N$. We have
$$\left|\frac 1 n-0\right|=\frac 1 n\leq \frac 1 N < \varepsilon.$$

That's the complete proof. Finding the proof is another matter. I will explain my process. Since the statement you want to prove is of the form "for all positive real numbers $\varepsilon$...", the proof should start with some version of the statement "Let $\varepsilon$ be an arbitrary positive real number". I think it's OK to simplify it to "Let $\varepsilon>0$".

So the first sentence of the proof writes itself. To continue, you need to figure out how to define your N. You want the inequality $\big|\frac 1 n -0\big|<\varepsilon$ to hold for all n such that $n\geq N$. That means in particular that you want the inequality to hold when n=N. So we need an N such that $\frac 1 N <\varepsilon$. This inequality is equivalent to $N>1/\varepsilon$. That explains my choice of N. The rest is straightforward.

Okay so I tried to prove it and I guess it went alright although it was missing some stuff. I have a question though, since we are dealing with a series that is converging, as we increase the numerical value of epsilon does the value of N go down? E.g. if Epsilon=0.5, N=10 and if Epsilon=1, N=2

 

[Fredrik]

Okay so I tried to prove it and I guess it went alright although it was missing some stuff. I have a question though, since we are dealing with a series that is converging, as we increase the numerical value of epsilon does the value of N go down? E.g. if Epsilon=0.5, N=10 and if Epsilon=1, N=2

Yes, $\varepsilon$ is half the length of the interval $(L-\varepsilon,L+\varepsilon)$, and N-1 is the number of the terms at the start of the sequence that aren't guaranteed to be in that interval. The smaller you make the interval, the more terms will end up outside it. For all $k$, if $x_k$ isn't in the interval, then N must be chosen greater than $k$.

It would be a good exercise for you to prove the theorems in post #16. If you post one of your proofs, we can discuss it here.

 

[Potatochip911]

Yes, $\varepsilon$ is half the length of the interval $(L-\varepsilon,L+\varepsilon)$, and N-1 is the number of the terms at the start of the sequence that aren't guaranteed to be in that interval. The smaller you make the interval, the more terms will end up outside it. For all $k$, if $x_k$ isn't in the interval, then N must be chosen greater than $k$.

It would be a good exercise for you to prove the theorems in post #16. If you post one of your proofs, we can discuss it here.

And I'm supposed to prove them using epsilon?

 

[Fredrik]

And I'm supposed to prove them using epsilon?

Yes, that's the only way, since this epsilon thing defines the meaning of the statements we want to prove.

When we write $\sum_{k=0}^\infty a_k=S$ we mean that S is a limit of the sequence $\big(\sum_{k=0}^n a_k\big)_{n=0}^\infty$ (the sequence of partial sums). And you know that this means that for all $\varepsilon>0$, there's a positive integer N such that the following implication holds for all positive integers n.
$$n\geq N\ \Rightarrow\ \left|\sum_{k=0}^n a_k-S\right|<\varepsilon.$$

Now suppose that $\sum_{k=0}^\infty b_k =S$ and that $N$ is a positive integer. What does the statement $\sum_{k=N+1}^\infty b_k=S-\sum_{k=0}^N b_k$ mean?

 

[Potatochip911]

Yes, that's the only way, since this epsilon thing defines the meaning of the statements we want to prove.

When we write $\sum_{k=0}^\infty a_k=S$ we mean that S is a limit of the sequence $\big(\sum_{k=0}^n a_k\big)_{n=0}^\infty$ (the sequence of partial sums). And you know that this means that for all $\varepsilon>0$, there's a positive integer N such that the following implication holds for all positive integers n.
$$n\geq N\ \Rightarrow\ \left|\sum_{k=0}^n a_k-S\right|<\varepsilon.$$

Now suppose that $\sum_{k=0}^\infty b_k =S$ and that $N$ is a positive integer. What does the statement $\sum_{k=N+1}^\infty b_k=S-\sum_{k=0}^N b_k$ mean?

I would think it means that the error is equal to the total value of the sum minus a finite sum. I am still kind of confused about this whole epsilon thing. Is it valid to say that $$|\sum_{n=0}^{N}a_{n}-S|=|\sum_{n=N+1}^{\infty}a_{n}|<\varepsilon$$ In the proofs for limits they also use $\delta$ for the domain restrictions, do we also want to do that here?

 

[Fredrik]

I would think it means that the error is equal to the total value of the sum minus a finite sum.

That's just literally what the equation says, in less precise language. To prove the theorem, you need to be able to explain the exact meaning of the statement, given by the definition of the notation and the definition of "limit". The exact meaning of that equality is that the real number on the right-hand side is the limit of the sequence of partial sums of the infinite series on the left-hand side. By definition of "limit", this means precisely that for all $\varepsilon>0$, there's a positive integer M such that the following implication holds for all positive integers m:
$$m\geq M\ \Rightarrow\ \left|\sum_{k=N+1}^m b_k - \left(S-\sum_{k=0}^N b_k\right)\right|<\varepsilon.$$ Now is this a true statement? If m>N, we can rewrite the inequality as $$\left|\sum_{k=0}^m b_k-S\right|<\varepsilon,$$ and this is certainly true when m is large enough. That's implied by the assumption $\sum_{k=0}^\infty b_k=S$.

Now let's turn these observations into a short and elegant proof of the theorem.

Theorem: If $\sum_{k=0}^\infty b_k$ is convergent and N is a positive integer, then $\sum_{k=0}^\infty b_k=\sum_{k=0}^N b_k +\sum_{k=N+1}^\infty b_k$.

Proof:
Let $\varepsilon>0$. Define $S=\sum_{k=0}^\infty b_k$. We will prove that $\sum_{k=N+1}^\infty b_k=S-\sum_{k=0}^N b_k.$ Let M be a positive integer such that the following implication holds for all positive integers m.
$$m\geq M\ \Rightarrow\ \left|\sum_{k=0}^m b_k-S\right|<\varepsilon.$$ (Such an M exists because $\sum_{k=0}^\infty b_k=S$). Let m be a positive integer such that $m\geq\max\{M,N+1\}$. We have
$$\left|\sum_{k=N+1}^m b_k - \left(S-\sum_{k=0}^N b_k\right)\right| =\left|\sum_{k=0}^m b_k-S\right|<\varepsilon.$$ Since $\varepsilon$ is an arbitrary positive real number and $m$ is an arbitrary positive integer such that $m\geq\max\{M,N+1\}$, this implies that $\sum_{k=N+1}^\infty b_k=S-\sum_{k=0}^N b_k.$

Is it valid to say that $$|\sum_{n=0}^{N}a_{n}-S|=|\sum_{n=N+1}^{\infty}a_{n}|<\varepsilon$$

Not without proof. The theorem I just proved tells us that the equality holds, and that the left-hand side is less than $\varepsilon$ if N is large enough. But we haven't assumed that N is large.

In the proofs for limits they also use $\delta$ for the domain restrictions, do we also want to do that here?

That $\delta$ only shows up when we're dealing with limits of functions. We're dealing with limits of sequences here, not limits of functions. Hence the slightly simpler definition, involving a real number N instead of a positive real number $\delta$.

 

[Ray Vickson]

That's just literally what the equation says, in less precise language. To prove the theorem, you need to be able to explain the exact meaning of the statement, given by the definition of the notation and the definition of "limit". The exact meaning of that equality is that the real number on the right-hand side is the limit of the sequence of partial sums of the infinite series on the left-hand side. By definition of "limit", this means precisely that for all $\varepsilon>0$, there's a positive integer M such that the following implication holds for all positive integers m:
$$m\geq M\ \Rightarrow\ \left|\sum_{k=N+1}^m b_k - \left(S-\sum_{k=0}^N b_k\right)\right|<\varepsilon.$$ Now is this a true statement? If m>N, we can rewrite the inequality as $$\left|\sum_{k=0}^m b_k-S\right|<\varepsilon,$$ and this is certainly true when m is large enough. That's implied by the assumption $\sum_{k=0}^\infty b_k=S$.

Now let's turn these observations into a short and elegant proof of the theorem.

Theorem: If $\sum_{k=0}^\infty b_k$ is convergent and N is a positive integer, then $\sum_{k=0}^\infty b_k=\sum_{k=0}^N b_k +\sum_{k=N+1}^\infty b_k$.

Proof:
Let $\varepsilon>0$. Define $S=\sum_{k=0}^\infty b_k$. We will prove that $\sum_{k=N+1}^\infty b_k=S-\sum_{k=0}^N b_k.$ Let M be a positive integer such that the following implication holds for all positive integers m.
$$m\geq M\ \Rightarrow\ \left|\sum_{k=0}^m b_k-S\right|<\varepsilon.$$ (Such an M exists because $\sum_{k=0}^\infty b_k=S$). Let m be a positive integer such that $m\geq\max\{M,N+1\}$. We have
$$\left|\sum_{k=N+1}^m b_k - \left(S-\sum_{k=0}^N b_k\right)\right| =\left|\sum_{k=0}^m b_k-S\right|<\varepsilon.$$ Since $\varepsilon$ is an arbitrary positive real number and $m$ is an arbitrary positive integer such that $m\geq\max\{M,N+1\}$, this implies that $\sum_{k=N+1}^\infty b_k=S-\sum_{k=0}^N b_k.$Not without proof. The theorem I just proved tells us that the equality holds, and that the left-hand side is less than $\varepsilon$ if N is large enough. But we haven't assumed that N is large.That $\delta$ only shows up when we're dealing with limits of functions. We're dealing with limits of sequences here, not limits of functions. Hence the slightly simpler definition, involving a real number N instead of a positive real number $\delta$.

This can be shortened a lot: If $S_k(x) = \sum_{j=0}^k a_j x^j$, then for $M > N$ we have
$$|S_M(x) - S_N(x)| \leq \sum_{j=N+1}^M |a_j| |x|^j = |a_{N+1}| \frac{|x|^{N+1}}{1-|x|} ( 1 - |x|^{M-N}) \leq |a_{N+1}| \frac{|x|^{N+1}}{1-|x|}$$
Calling that last quantity $r_N(x)$, we have
$$-r_N(x) \leq S_M(x) - S_N(x) \leq r_N(x)$$
Taking the limit as $M \to \infty$ we have $-r_N(x) \leq S(x) - S_N(x) \leq r_N(x)$, which is what the OP wants.

 

[Fredrik]

This can be shortened a lot: If $S_k(x) = \sum_{j=0}^k a_j x^j$, then for $M > N$ we have
$$|S_M(x) - S_N(x)| \leq \sum_{j=N+1}^M |a_j| |x|^j = |a_{N+1}| \frac{|x|^{N+1}}{1-|x|} ( 1 - |x|^{M-N}) \leq |a_{N+1}| \frac{|x|^{N+1}}{1-|x|}$$
Calling that last quantity $r_N(x)$, we have
$$-r_N(x) \leq S_M(x) - S_N(x) \leq r_N(x)$$
Taking the limit as $M \to \infty$ we have $-r_N(x) \leq S(x) - S_N(x) \leq r_N(x)$, which is what the OP wants.

This is a proof of an entirely different theorem, not a shorter version of my proof. You're just showing him a solution to the problem he asked about in post #1.

Your solution is mathematically sound (just like the OP's own solution, when the elementary error in post #1 is corrected), but (just like the OP's own solution) it requires the reader to understand a few things other than the basic definitions, that are elementary to someone who has a lot of experience with these proofs, but are difficult to someone who's just getting started. Your proof requires the reader to understand that the sequence $\big(\big|\sum_{j=0}^M a_j x_j-\sum_{j=0}^N a_j x_j\big|\big)_{M=N+1}^\infty$ converges to $\big|\sum_{j=0}^\infty a_j x_j-\sum_{j=0}^N a_j x_j\big|$, that it's an increasing sequence, and that the limit of an increasing sequence is the least upper bound of the set of terms.

My solution, which I haven't posted yet, is a more straightforward application of the definitions. It requires the reader to understand limits of sequences and what it means to say that a series is convergent, but not much more than that.

 

[Ray Vickson]

This is a proof of an entirely different theorem, not a shorter version of my proof. You're just showing him a solution to the problem he asked about in post #1.

Your solution is mathematically sound (just like the OP's own solution, when the elementary error in post #1 is corrected), but (just like the OP's own solution) it requires the reader to understand a few things other than the basic definitions that are elementary to someone who has a lot of experience with these proofs, but are difficult to someone who's just getting started. Your proof requires the reader to understand that the sequence $\big(\big|\sum_{j=0}^M a_j x_j-\sum_{j=0}^N a_j x_j\big|\big)_{M=N+1}^\infty$ converges to $\big|\sum_{j=0}^\infty a_j x_j-\sum_{j=0}^N a_j x_j\big|$, that it's an increasing sequence, and that the limit of an increasing sequence is the least upper bound of the set of terms.

My solution, which I haven't posted yet, is a more straightforward application of the definitions. It requires the reader to understand limits of sequences and what it means to say that a series is convergent, but not much more than that.

You say that I (essentially) required the reader to understand that $|S_M - S_N| \to |S-S_N|$ and that it's an increasing sequence. Not so: I re-wrote $|S_M - S_N| \leq r_N$ as $-r_N \leq S_M - S_N \leq r_N$ precisely to get rid of the absolute-value signs---basically, to avoid the very issue you raise. Then, the only properties used after that were that (i) that $S = \lim_M S_M$ by definition of $S$ (assumed to exist in the first place); (ii) that $U \leq T_M \leq V \Rightarrow U \leq \lim_M T_M \leq V$; and (iii) $\lim_M (S_M - c) = (\lim_M S_M) - c$, where $c = S_N$ is a constant.

 

[Fredrik]

You say that I (essentially) required the reader to understand that $|S_M - S_N| \to |S-S_N|$ and that it's an increasing sequence. Not so: I re-wrote $|S_M - S_N| \leq r_N$ as $-r_N \leq S_M - S_N \leq r_N$ precisely to get rid of the absolute-value signs---basically, to avoid the very issue you raise. Then, the only properties used after that were that (i) that $S = \lim_M S_M$ by definition of $S$ (assumed to exist in the first place); (ii) that $U \leq T_M \leq V \Rightarrow U \leq \lim_M T_M \leq V$; and (iii) $\lim_M (S_M - c) = (\lim_M S_M) - c$, where $c = S_N$ is a constant.

OK, I agree that this way of explaining what you did is better. My way of explaining it would have been been better if you hadn't rewritten the inequalities without the absolute value. I didn't see the point of that rewrite before, but I see now that it makes the first step of your explanation trivial, while the first step of mine isn't.

 

[Fredrik]

@Potatochip911, can you show us any proof at all that involves the definition of limit and that hasn't yet been posted in this thread? I suggest that you try to prove one of the following. Actually, you should give them both a try, but if you think you have proved both, it's probably better if you only post one of the proofs, so that we can discuss it before you type up the other.

Theorem 1: If $(S_n)_{n=1}^\infty$ is convergent and c is a real number, then $(S_k-c)_{n=1}^\infty$ is convergent, and we have
$$\lim_n(S_n-c)=\lim_n S_n-c$$.

Theorem 2: If $\sum_{k=1}^\infty b_k$ is convergent and c is a real number, then $\sum_{k=1}^\infty cb_k$ is convergent, and we have
$$\sum_{k=1}^\infty cb_k =c\sum_{k=1}^\infty b_k.$$
These are as easy to prove as any theorem that involves the definition of "limit" (of a sequence). So I suggest that you try to prove them both and post one of your proofs here.

Edit: Another good exercise for you (that doesn't involve the definition of limit, but does involve an arbitrary positive real number, just like that definition) is to prove the following:

Theorem 3: If $x<y+\varepsilon$ for all $\varepsilon>0$, then $x\leq y$.

This isn't just another exercise. I'm using this result in my solution of the problem you posted. Hint: Use an indirect proof, i.e. prove that if the statement on the right is false, then so is the statement on the left.

 

[Potatochip911]

@Potatochip911, can you show us any proof at all that involves the definition of limit and that hasn't yet been posted in this thread? I suggest that you try to prove one of the following. Actually, you should give them both a try, but if you think you have proved both, it's probably better if you only post one of the proofs, so that we can discuss it before you type up the other.

Theorem 1: If $(S_n)_{n=1}^\infty$ is convergent and c is a real number, then $(S_k-c)_{n=1}^\infty$ is convergent, and we have
$$\lim_n(S_n-c)=\lim_n S_n-c$$.

Theorem 2: If $\sum_{k=1}^\infty b_k$ is convergent and c is a real number, then $\sum_{k=1}^\infty cb_k$ is convergent, and we have
$$\sum_{k=1}^\infty cb_k =c\sum_{k=1}^\infty b_k.$$
These are as easy to prove as any theorem that involves the definition of "limit" (of a sequence). So I suggest that you try to prove them both and post one of your proofs here.

Edit: Another good exercise for you (that doesn't involve the definition of limit, but does involve an arbitrary positive real number, just like that definition) is to prove the following:

Theorem 3: If $x<y+\varepsilon$ for all $\varepsilon>0$, then $x\leq y$.

This isn't just another exercise. I'm using this result in my solution of the problem you posted. Hint: Use an indirect proof, i.e. prove that if the statement on the right is false, then so is the statement on the left.

I didn't understand how to prove the first one so I decided to skip it. For the second one: $$\sum_{k=1}^{\infty} b_{k}=S; \space \space \space \mbox{Let } m \geq M \space \space \& \space \space \varepsilon > 0\\
\Longrightarrow |\sum_{k=1}^{m}-S|<\varepsilon
$$ I'm pretty sure it's also valid to write down
$$c \in \mathbb{R} \space \space |\sum_{k=1}^{m}cb_{k}-cS| < \varepsilon \\
|c(\sum_{k=1}^{m}b_{k}-S)|<\varepsilon \\
|\sum_{k=1}^{m}b_{k}-S|<\frac{\varepsilon}{|c|}
$$
I was also thinking that maybe we can just multiply both sides to get $c|\sum_{k=1}^{m}b_{k}-S|<c\varepsilon \Rightarrow |\sum_{k=1}^{m}cb_{k}-cS|<c\varepsilon$ and since $c$ is a real number I would also like to assume that $\varepsilon$ is real and therefore multiplying results in a finite number. I'm thinking that if it multiplying the sequence by $c$ made the sequence diverge we would be left with some sort of indeterminate form on one side.
I'm not quite sure this proves it but I figured I would at least try this. Also I don't understand why we specify that $m \geq M$ since we never use $M$ in our calculations.

 

[Fredrik]

$$\sum_{k=1}^{\infty} b_{k}=S; \space \space \space \mbox{Let } m \geq M \space \space \& \space \space \varepsilon > 0\\
\Longrightarrow |\sum_{k=1}^{m}-S|<\varepsilon$$

It's going to be useful to have a symbol for the sum $\sum_{k=1}^\infty b_k$, so I like that you assigned the symbol S right away.

When you say "let $m\geq M$", this is interpreted as "let m be an arbitrary integer such that $m\geq M$". This isn't a meaningful thing to say unless you have already specified what M is. If you want both m and M to be arbitrary apart from the fact that they satisfy that inequality, you have to say something like "let m and M be integers such that $m\geq M$". (This would be a bad idea in this proof).

It's correct to start with "let $\varepsilon>0$", but you can't leave M arbitrary. You need to let M be a positive integer such that the implication
$$m\geq M\ \Rightarrow\ \left|\sum_{k=1}^m b_k-S\right|<\frac{\varepsilon}{|c|}$$
holds for all positive integers m such that $m\geq M$. The existence of such an M is implied by the definition of the statement $\sum_{k=1}^\infty b_k=S$. The reason why we're saying that M is such an integer is that we're preparing to use the assumption that $\sum_{k=1}^\infty b_k=S$. Why the |c| in the denominator? The reason will become clear when we get to the end of the proof. For now, just note that the definition of limit allows us to put that |c| there.

I'm pretty sure it's also valid to write down
$$c \in \mathbb{R} \space \space |\sum_{k=1}^{m}cb_{k}-cS| < \varepsilon $$

This looks a lot like the statement you want to prove, and it's never OK to just write that down as if it's known to be a true statement. Also, since you have left M unspecified, the condition $m\geq M$ may not be sufficient to ensure that this inequality holds.

What you need to do is to let $m$ be an (arbitrary) integer such that $m\geq M$ (in more difficult proofs, you would have to replace M with a different number at this point*), write down only the left-hand side of the inequality you want to prove, i.e. $\big|\sum_{k=1}^m cb_k- cS\big|$, and then work with that expression (using what we said at the start of the proof), until you have found the inequality we want, i.e.
$$\left|\sum_{k=1}^m cb_k- cS\right|<\varepsilon.$$

*) For an example, see the proof in post #24, where I used $\max\{M,N+1\}$ instead of $M$.

 

[Potatochip911]

This looks a lot like the statement you want to prove, and it's never OK to just write that down as if it's known to be a true statement. Also, since you have left M unspecified, the condition $m\geq M$ may not be sufficient to ensure that this inequality holds.

Okay I am slightly confused now since I was able to get $|\sum_{k=1}^{m}b_{k}-S|<\frac{\varepsilon}{|c|}$ by starting with $|\sum_{k=1}^{m}cb_{k}-cS|<\varepsilon$, if I can't assume this is true I'm not sure how to end up with $|c|$ in the denominator.

 

[Fredrik]

Okay I am slightly confused now since I was able to get $|\sum_{k=1}^{m}b_{k}-S|<\frac{\varepsilon}{|c|}$ by starting with $|\sum_{k=1}^{m}cb_{k}-cS|<\varepsilon$, if I can't assume this is true I'm not sure how to end up with $|c|$ in the denominator.

You're not supposed to end up with $|\sum_{k=1}^{m}b_{k}-S|<\frac{\varepsilon}{|c|}$. You simply define M as a positive integer such that this inequality holds for all $m\geq M$. Then you use that inequality to prove that $|\sum_{k=1}^{m}cb_{k}-cS|<\varepsilon$ for all $m$ such that $m\geq M$.

Recall that the statement $\sum_{k=1}^\infty b_k=S$ means that for all $\varepsilon>0$, there's a positive integer M such that the following implication holds for all positive integers m. $$m\geq M\ \Rightarrow\ \left|\sum_{k=1}^m b_k-S\right|<\varepsilon.$$ The "for all" is the key. If you let $\varepsilon$ be an arbitrary positive real number, then $\varepsilon/|c|$ is a positive real number too. So the statement $\sum_{k=1}^\infty b_k=S$ implies that there's a positive integer M such that the following implication holds for all positive integers m.
$$m\geq M\ \Rightarrow\ \left|\sum_{k=1}^m b_k-S\right|<\frac{\varepsilon}{|c|}.$$

 

[Potatochip911]

You're not supposed to end up with $|\sum_{k=1}^{m}b_{k}-S|<\frac{\varepsilon}{|c|}$. You simply define M as a positive integer such that this inequality holds for all $m\geq M$. Then you use that inequality to prove that $|\sum_{k=1}^{m}cb_{k}-cS|<\varepsilon$ for all $m$ such that $m\geq M$.

Recall that the statement $\sum_{k=1}^\infty b_k=S$ means that for all $\varepsilon>0$, there's a positive integer M such that the following implication holds for all positive integers m. $$m\geq M\ \Rightarrow\ \left|\sum_{k=1}^m b_k-S\right|<\varepsilon.$$ The "for all" is the key. If you let $\varepsilon$ be an arbitrary positive real number, then $\varepsilon/|c|$ is a positive real number too. So the statement $\sum_{k=1}^\infty b_k=S$ implies that there's a positive integer M such that the following implication holds for all positive integers m.
$$m\geq M\ \Rightarrow\ \left|\sum_{k=1}^m b_k-S\right|<\frac{\varepsilon}{|c|}.$$

So to prove this I want to assume that for all $m\geq M \Rightarrow |\sum_{k=1}^{m}b_{k}-S|<\varepsilon$ with $M\in\mathbb{N}, M\neq0$ and now it's also valid to assume this holds? $$m\geq M \Rightarrow |\sum_{k=1}^{m}b_{k}-S|<\frac{\varepsilon}{|c|} \space \space \mbox{which leads to} \\
|c||\sum_{k=1}^{m}b_{k}-S|<\varepsilon \\
|c\sum_{k=1}^{m}b_{k}-cS|<\varepsilon \\
|\sum_{k=1}^{m}cb_{k}-cS|<\varepsilon $$

 

[Fredrik]

So to prove this I want to assume that for all $m\geq M \Rightarrow |\sum_{k=1}^{m}b_{k}-S|<\varepsilon$

It's a bit odd to use the "for all" and the implication arrow like that. It would be better to say e.g. "for all $m\geq M$, we have $|\sum_{k=1}^{m}b_{k}-S|<\varepsilon$" or "for all $m$, we have $m\geq M \Rightarrow |\sum_{k=1}^{m}b_{k}-S|<\varepsilon$".

We don't assume that this implication holds. Our assumption is that $\sum_{k=1}^\infty b_k$ is convergent. If we denote the sum of the series by $S$, then this assumption implies that there's an M such that the implication holds for all m.

and now it's also valid to assume this holds? $$m\geq M \Rightarrow |\sum_{k=1}^{m}b_{k}-S|<\frac{\varepsilon}{|c|} \space \space \mbox{which leads to} \\
|c||\sum_{k=1}^{m}b_{k}-S|<\varepsilon \\
|c\sum_{k=1}^{m}b_{k}-cS|<\varepsilon \\
|\sum_{k=1}^{m}cb_{k}-cS|<\varepsilon $$

The first line is the assumption. The last line is a consequence of that assumption.

OK, let's clean this up a little. Here's how I would type the proof:

Let $\varepsilon>0$. Let M be a positive integer such that the following implication holds for all positive integers m.
$$m\geq M\ \Rightarrow\ \left|\sum_{k=1}^m b_k-S\right|<\frac{\varepsilon}{|c|}.$$ For all $m\geq M$, we have
$$\left|\sum_{k=1}^m cb_k-cS\right| = |c|\left|\sum_{k=1}^m b_k -S\right|<|c|\frac{\varepsilon}{|c|}=\varepsilon.$$

I think the best way to approach proofs like this is to start by looking at the statement you want to prove. (This is when you're trying to find the proof). That statement is $\sum_{k=1}^\infty cb_k=cS$. It means that for all $\varepsilon>0$, there's a positive integer M such that the following implication holds for all positive integers m.
$$m\geq M\ \Rightarrow\ \left|\sum_{k=1}^m cb_k-cS\right| <\varepsilon.$$ Because of the "for all" at the start of that statement, the proof should start with "let $\varepsilon>0$". Now the question is how to choose M. Since the inequality that's part of the the statement that we want to prove is equivalent to $\big|\sum_{k=1}^m b_k-S\big|<\frac{\varepsilon}{|c|}$, all we have to do is to choose M such that this inequality holds for all m such that $m\geq M$. When you type up the proof, you start with this choice of M, and then use it to prove the statement we want to prove.