# Rearangement of series

1. Jun 8, 2007

### quasar987

My book proves that if a series is absolutely convergent, then every rearangement is absolutely convergent also.

They then argue that the thm does not hold for conditionally convergeant series and give as a counter exemple the following thing: Let $\sum a_n$ be a conditionally convergent series and define the positive part of {a_n} by p_n = a_n if a_n > 0 and =0 otherwise and the negative part of {a_n} by q_n = a_n if a_n < 0 and =0 otherwise. Then observe that

$$p_n=\frac{a_n+|a_n|}{2}$$

and

$$q_n=\frac{a_n-|a_n|}{2}$$

or that inversely,

$$|a_n|=p_n-q_n$$ and $$a_n=p_n+q_n$$

But what does this prove? Where is the rearangement?

2. Jun 8, 2007

### ice109

all i can imagine is that neither of those new smaller series is convergent

edit

i think only $$p_n$$ is divergent so i don't know what that does

mathworld says

If $$\Sigma u_k$$ and $$\Sigma v_k$$ are convergent series, then $$\Sigma (u_k+v_k)$$ and $$\Sigma (u_k-v_k)$$ are convergent.

but i don't know if the reverse ( or w/e the proper word is ) has to be true

where if $$\Sigma u_k$$ is divergent then $$\Sigma (u_k+v_k)$$ is also divergent

Last edited: Jun 8, 2007
3. Jun 9, 2007

### quasar987

Ugh. Sorry ice109 if you spent time thinking about this. I just realized that the definitions and observations made in my OP are just a kind of lemma to Riemann's theorem that every conditionally convergent series can be rearranged to converge to any real number or to diverge.

4. Jun 9, 2007

### ice109

which theorem is that?

5. Jun 9, 2007

### quasar987

well I just stated it:

"every conditionally convergent series can be rearranged to converge to any real number or to diverge."

6. Jun 9, 2007

### ice109

i meant which theorem so i coud look up a proof

7. Jun 9, 2007

### quasar987

What more can I say other than that it's a thm of Riemann and give you the exact statement?