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April 1st, 2017, 01:40 PM   #1
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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.
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April 1st, 2017, 04:14 PM   #2
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You are changing your rearrangement after the point that the proof specifies.
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April 1st, 2017, 09:11 PM   #3
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Quote:
Originally Posted by v8archie View Post
You are changing your rearrangement after the point that the proof specifies.
That bothered me.

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.
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April 2nd, 2017, 12:06 AM   #4
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Quote:
Originally Posted by zylo View Post
...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?
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.
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April 3rd, 2017, 09:46 AM   #5
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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
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April 3rd, 2017, 10:06 AM   #6
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Quote:
Originally Posted by zylo View Post

Let S'=1+3+5+.......+2+4+6+........
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.
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April 3rd, 2017, 05:56 PM   #7
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It's an interesting argument, but flawed.
Quote:
Originally Posted by zylo View Post
No matter how large m is, S'm will not contain 4.
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.
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April 4th, 2017, 04:21 AM   #8
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Quote:
Originally Posted by v8archie View Post
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.
Every element in my series has a natural number subscript. It's just that the odd subscripts come first.

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.
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April 4th, 2017, 04:50 AM   #9
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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.
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April 4th, 2017, 05:21 AM   #10
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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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