Absolutely Convergent Series Rearangement Proof: Counterexample and Explanation

  • Thread starter Thread starter quasar987
  • Start date Start date
  • Tags Tags
    Series
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
6 replies · 2K views
Science Advisor
Homework Helper
Gold Member
Messages
4,796
Reaction score
32
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 [itex]\sum a_n[/itex] 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

[tex]p_n=\frac{a_n+|a_n|}{2}[/tex]

and

[tex]q_n=\frac{a_n-|a_n|}{2}[/tex]

or that inversely,

[tex]|a_n|=p_n-q_n[/tex] and [tex]a_n=p_n+q_n[/tex]

But what does this prove? Where is the rearangement?
 
Physics news on Phys.org
quasar987 said:
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 [itex]\sum a_n[/itex] 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

[tex]p_n=\frac{a_n+|a_n|}{2}[/tex]

and

[tex]q_n=\frac{a_n-|a_n|}{2}[/tex]

or that inversely,

[tex]|a_n|=p_n-q_n[/tex] and [tex]a_n=p_n+q_n[/tex]

But what does this prove? Where is the rearangement?

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

edit

i think only [tex]p_n[/tex] is divergent so i don't know what that does

mathworld says

If [tex]\Sigma u_k[/tex] and [tex]\Sigma v_k[/tex] are convergent series, then [tex]\Sigma (u_k+v_k)[/tex] and [tex]\Sigma (u_k-v_k)[/tex] are convergent.

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

where if [tex]\Sigma u_k[/tex] is divergent then [tex]\Sigma (u_k+v_k)[/tex] is also divergent
 
Last edited:
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.
 
quasar987 said:
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.

which theorem is that?
 
well I just stated it:

"every conditionally convergent series can be rearranged to converge to any real number or to diverge."
 
quasar987 said:
well I just stated it:

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

i meant which theorem so i could look up a proof
 
What more can I say other than that it's a thm of Riemann and give you the exact statement? :confused: