# Prove Sum Approximation Theorem

1. Jun 17, 2015

### Potatochip911

1. The problem statement, all variables and given/known data

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

2. Relevant equations
3. The attempt at a solution

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

2. Jun 17, 2015

### Fredrik

Staff Emeritus
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 $|S-\sum_{n=0}^N a_n x^n|$.

3. Jun 17, 2015

### Potatochip911

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

4. Jun 17, 2015

### Dick

$\frac{1}{1-|x|}$ 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?

5. Jun 17, 2015

### Fredrik

Staff Emeritus
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:
$$m\geq M\ \Rightarrow\ \left|\sum_{n=0}^m a_n x^n-S\right|<\varepsilon.$$ 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.

Last edited: Jun 18, 2015
6. Jun 17, 2015

### Potatochip911

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

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.

7. Jun 17, 2015

### Fredrik

Staff Emeritus
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.

8. Jun 17, 2015

### Potatochip911

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

9. Jun 18, 2015

### Fredrik

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

Last edited: Jun 18, 2015
10. Jun 18, 2015

### 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*

11. Jun 18, 2015

### Potatochip911

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.

12. Jun 18, 2015

### Fredrik

Staff Emeritus
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).

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 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 $\big|\sum_{i=1}^n x_i\big|\leq\sum_{i=1}^n|x_i|.$

Last edited: Jun 19, 2015
13. Jun 18, 2015

### ChrisVer

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.

14. Jun 18, 2015

### Fredrik

Staff Emeritus
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.

Last edited: Jun 18, 2015
15. Jun 20, 2015

### 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.

16. Jun 20, 2015

### Fredrik

Staff Emeritus
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.

17. Jun 20, 2015

### Potatochip911

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.

18. Jun 20, 2015

### Fredrik

Staff Emeritus
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:

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.

19. Jun 20, 2015

### Potatochip911

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

20. Jun 20, 2015

### Fredrik

Staff Emeritus
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.