April 1st, 2017, 01:40 PM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,329 Thanks: 94  Rearrangement of Series
Rearrangement of Series Let $\displaystyle \Sigma$an be absolutely convergent. A proof that any rearrangement is absolutely convergent and has the same sum contains the following: Let $\displaystyle \Sigma$bn be any rearrangement of $\displaystyle \Sigma$an. Let S'm be the mth partial sum of $\displaystyle \Sigma$bn. For sufficiently large m, S'm will contain a1,a2,...an. Let $\displaystyle \Sigma$bn =a2+a3+...am+a1+...., no matter what m is. S'm can never contain a1. 
April 1st, 2017, 04:14 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,268 Thanks: 2434 Math Focus: Mainly analysis and algebra 
You are changing your rearrangement after the point that the proof specifies.

April 1st, 2017, 09:11 PM  #3  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,329 Thanks: 94  Quote:
Let S'm = a2+a3+...am + a1 ie, the process of checking changes (creates) the sequence. Let me see the first term of your sequence: see, no a1 Let me see two terms of your sequence: see, no a1 ... ... Just thinking out loud. But you are right. In any actual rearrangement you can get to a1, unless a1 is at the "end" of the sequence, where the "let me see" procedure above puts it. EDIT: On the other hand, if we are talking any endless rearrangement, what guarantee is there that you will get to a1? Isn't there something that says in a sequence of numbers (subscripts), you will eventually get to a specific number if it's in the sequence? Last edited by zylo; April 1st, 2017 at 09:35 PM.  
April 2nd, 2017, 12:06 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,268 Thanks: 2434 Math Focus: Mainly analysis and algebra  Every element in the series is in a position denoted by a finite ordinal. So you will eventually reach any particular element that you are looking for. By extension, given any finite set of elements, you will also eventually reach a point where you have seen every one of them.

April 3rd, 2017, 09:46 AM  #5 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,329 Thanks: 94 
Any infinite sequence can be broken up into infinite sub sequences. For example: (1,2,3,4,5,...)=(1,3,5,...) "+" (2,4,6,....) Let S=1+2+3+....... and Sm=sum of m terms. Let S'=1+3+5+.......+2+4+6+........ and S'm = sum of m terms No matter how large m is, S'm will not contain 4. {Just to save writing, I have associated n with an an or bn.) no 
April 3rd, 2017, 10:06 AM  #6 
Senior Member Joined: Aug 2012 Posts: 1,850 Thanks: 509  This is not a sequence or a series. It has order type $\omega + \omega$. A sequence (or a series, if you put plus signs between the terms) must have order type $\omega$. It's true that you have a bijection between your original sequence and your rearranged one, but it's no longer a sequence (or series). To see that, note that both $1$ and $2$ have no immediate predecessor. In a sequence, only the first item has no immediate predecessor.

April 3rd, 2017, 05:56 PM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,268 Thanks: 2434 Math Focus: Mainly analysis and algebra 
It's an interesting argument, but flawed. If there is no partial sum that contains the number 4, then 4 is not in the sequence/series. Remember that the elements of the series are indexed by the natural numbers: every element has a natural number subscript. Your "arrangement" of terms doesn't do this. 
April 4th, 2017, 04:21 AM  #8  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,329 Thanks: 94  Quote:
The rearrangement of absolutely convergent series theorem allows any rearrangement. By extension, any absolutely convergent series can be written as the sum of any number of absolutely convergent series. The question is, would the sum always be the same? Last edited by zylo; April 4th, 2017 at 04:35 AM.  
April 4th, 2017, 04:50 AM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,268 Thanks: 2434 Math Focus: Mainly analysis and algebra 
What is the subscript of 2 then? Hint: it doesn't have a natural number subscript because the odd numbers exhaust all of the natural number subscripts. Last edited by v8archie; April 4th, 2017 at 04:56 AM. 
April 4th, 2017, 05:21 AM  #10 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,329 Thanks: 94 
2 doesn't have a subscript. But I see your point: a1+a2+a3+a4+...=a1+a3+.......+a2+a4+......=b1+b2+b 3+b4+.... All the natural numbers appear as suffixes in both sums but I can't use the standard proof which requires a partial sum S'm for which I can't always specify m. Not saying the theorem is false, just noting a possible flaw in the proof. 

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