# Arrangement Of Terms Question

1. Aug 4, 2009

### JG89

Suppose the infinite series $$\sum a_v$$ is NOT absolutely convergent. Suppose it also has an infinite amount of positive and an infinite amount of negative terms.

Say we want to prove it converges by proving the sequence of partial sums, A_n, are Cauchy.

Then we need to prove that for every positive epsilon, $$|A_n - A_m| < \epsilon$$, for n and m sufficiently large.

Note that $$|A_n - A_m| = |a_1 + a_2 + ... + a_n - (a_1 + a_2 + ... + a_m)| = |a_{n+1} + a_{n+2} + ... + a_m|$$.

Some of these terms are positive, some negative. Are we allowed to "separate" the positive and negative terms? Like this:

Say P_n is the sequence of positive terms and N_n is the sequence of negative terms in the sum $$|a_{n+1} + a_{n+2} + ... + a_m|$$. Then can we write $$|a_{n+1} + a_{n+2} + ... + a_m| = |P_1 + P_2 + ... + P_i + N_1 + N_2 + ... + N_j|$$ ?

It seems to me like this should be fine, since this is a finite sum. But we are going to have to take n and m larger and larger for epsilon getting smaller and smaller, so I am not sure.

Last edited: Aug 4, 2009
2. Aug 6, 2009

### Billy Bob

It's fine. Epsilon is very small, but fixed, since your first step will be "let epsilon be a fixed but arbitrary positive number." Later, m and n will be very large, but still finite.

3. Aug 6, 2009

### Elucidus

A sum of a finite number of terms can always be rearranged owing to commutativity of addition. So in your example, provided m and n are finite, you can separate the summands into positive and negative terms. If m or n are allowed to wander to infinity, though, things get stickier.

Commutativity fails in a sum of an infinite number of terms if the sum is only conditional convergent. Riemann's Rearrangement Theorem says that for any conditionally convergent series, there exists a rearrangement of the terms were the sum diverges to infinity, negative infinity, or any arbitrary constant.

I hope this is helpful.

--Elucidus

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