# Homework Help: Reconciling Reimann's Series Theorem

1. Oct 12, 2014

### eyesontheball1

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

It's well known that Reimann's Rearrangement Theorem states that if a real-valued series is conditionally convergent, then its terms can be rearranged to sum to any arbitrary real number. I've arrived at a result that doesn't agree with this statement, and so I'd greatly appreciate it if someone could explain to me why my result differs from what's stated in the said theorem. I think I might know why my result differs, but I'd be grateful if someone else could confirm for me why my result differs if possible. Thanks in advance!

p.s. - I'm sorry for not typing out everything below using LateX code. It's been a while since I've written LateX code and this entire post would have taken me much longer if I were to take the time to brush up on LateX syntax. I know this makes it more of a hassle for someone to read through and comprehend my post, and so I apologize for this.

2. Relevant equations

S_n = sum[k=1, n]{a_k}, where all the a_k are real-valued

S = sum[k=1, infinity]{a_k}

S_n(sigma) = sum[k=1, n]{a_sigma(k)}, where sigma: First n positive integers -> First n positive integers, so that sigma is a bijection.

S(sigma) = sum[k=1, infinity]{a_sigma(k)}, where sigma: The positive integers -> The positive integers, so that sigma is a bijection.

3. The attempt at a solution
When n=2, we have S_2 = sum[k=1, 2]{a_k} = sum[k=1, 2]{a_sigma(k)} = S_2(sigma), where sigma is any bijection from {1,2} to {1,2}.

Assuming the equation, S_n = sum[k=1, n]{a_k} = sum[k=1, n]{a_sigma(k)} = S_n(sigma), with sigma being any bijection from the first n positive integers to the first n positive integers, holds for some arbitrary positive integer n that's greater than 2, we can prove that the equation, S_n+1 = S_n+1(sigma), with sigma being any bijection from the first n+1 positive integers to the first n+1 positive integers, also holds, concluding by induction that the equation, S_N = S_N(sigma), holds for any arbitrary positive integer N and for any bijection between the first N positive integers as follows:

S_n = S_n(sigma) => sum[k=1, n]{a_k} = sum[k=1, n]{a_sigma(k)} =>

sum[k=1, n]{a_k} + a_k+1 = sum[k=1, n]{a_sigma(k)} + a_k+1 =>

S_n+1 = sum[k=1, n]{a_sigma(k)} + a_k+1,

and sum[k=1, n]{a_sigma(k)} + a_k+1 = sum[k=1, sigma(k) =/ n+1 for all k /= n+1, n+1]{a_sigma(k)}

= sum[k=1, n+1]{a_sigma(k)} = S_n+1(sigma) =>

S_n+1 = S_n+1(sigma).

Therefore, the equation S_n = S_n(sigma) holds for any positive integer n and any possible bijection, sigma, between the first n positive integers.

With this in mind, let's consider the sum, S'_n(sigma) = sum[k=1, n]{a'_sigma(k)} = sum[k=1, n]{a_k - a_sigma(k)} = S_n - S_n(sigma).

We know that for any positive integer, n, we must have S'_n(sigma) = 0.

It then follows that for all epsilon > 0, there exists a positive integer, N, s.t. if n > N, then 0 <= | S'_n(sigma) - 0 | < epsilon, since, regardless of the value of epsilon that we choose, any positive integer N that we choose will suffice to make this statement true. Hence, we have that...

{For all epsilon > 0, there exists a positive integer, N, s.t. n > N => 0 <= | S'_n(sigma) - 0 | < epsilon} =>

lim[n -> infinity]{S'_n(sigma)} = 0 =>

lim[n -> infinity]{S_n - S_n(sigma)} = 0 =>

lim[n -> infinity]{S_n} = lim[n -> infinity]{S_n(sigma)} =>

S = S(sigma), where sigma is any bijection between the positive integers.

2. Oct 12, 2014

### RUber

The concern I have with this is that your bijection sigma is always limited to whatever finite n you are working with. You are essentially concluding that any finite sum is equal to the rearrangement of the finite sum and the rest of the series is identical at any n. If you were to allow sigma to be any bijection from $\mathbb{Z}^+ \to \mathbb{Z}^+$ you would not be able to make the same claims about equality in the finite sums, though it would be expected that the infinite sum would be the same. This, I think, is the heart of the rearrangement theorem. You can have different limits to the finite sums which would lead to concluding that the infinite sum could be equal to multiple values. In your example, you are fixing the finite sums to have the same values, and implicitly stating that the infinite tails are identical, since the bijection of the first n integers would not apply to the tails.

3. Oct 12, 2014

### eyesontheball1

That was my concern as well. I'm a little confused by the part of your response where you say "If you were to allow sigma to be any bijection from Z+→Z+ you would not be able to make the same claims about equality in the finite sums, though it would be expected that the infinite sum would be the same. This, I think, is the heart of the rearrangement theorem. You can have different limits to the finite sums which would lead to concluding that the infinite sum could be equal to multiple values.". At first you state that if I were to allow sigma to be any bijection, then the finite sums would not necessarily be equal, although the infinite sum would be expected to be the same, and then you state that the different limits to the finite sums would lead to an infinite sum that could be equal to multiple values. When you use the word 'expected' here, you're not stating that the infinite sum actually would be the same, right? In other words, you're using the word 'expected' in the sense that intuition would lead one to believe that the infinite sums would be the same, although the truth is the contrary, right?

4. Oct 12, 2014

### gopher_p

The following are provable theorems:

1. If $\sum a_n$ is convergent but not absolutely convergent, then for all $r\in\mathbb{R}$, there is a permutation $\sigma$ of $\mathbb{N}$ such that $\sum a_{\sigma(n)}=r$.

2. If $\sum a_n$ is convergent and $\sigma$ is a permutation of $\mathbb{N}$ such that there is $M\in\mathbb{N}$ with $|\sigma(n)-n|\leq M$ for all $n$, then $\sum a_{\sigma(n)}$ is convergent and $\sum a_{\sigma(n)}=\sum a_n$.

Your result relies on permutations that are "bounded" like those in (2).

5. Oct 12, 2014

### RUber

That's right. Taking the infinite sum to be the limit of finite sums allows us to define it to be a variety of values based on choice of arrangement. Although the terms in either arrangement of infinite sums is the same, so the definition of the sum should intuitively be the same.

6. Oct 12, 2014

### eyesontheball1

Gotcha. Thanks, guys!