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## Homework Statement

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.[/B]

## Homework 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.

## 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.