Unconditional Convergence

chiph588@

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Is there a series that's unconditionally convergent but not absolutely convergent?
 

Drexel28

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Is there a series that's unconditionally convergent but not absolutely convergent?
Isn't that the defintion. A series is conditionally if it's convergent but it's modulus isn't. So, if it's unconditionally convergent either (I would assume) that it doesn't converge at all or it's absolutely convergent.
 
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chiph588@

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Sep 2008
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Champaign, Illinois
Isn't that the defintion. A series is conditionally if it's convergent but it's modulus isn't. So, if it's unconditionally convergent either (I would assume) that it doesn't converge at all or it's absolutely convergent.
That's what I thought, but I've only seen that absolutely convergent implies unconditional convergence.
 

chiph588@

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Wait a minute: "Absolute convergence and convergence together imply unconditional convergence, but unconditional convergence does not imply absolute convergence in general".

This is from Wikipiedia.

I'm really curious to see a series with this property.
 

Drexel28

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Mar 2010
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Bratislava
Is there a series that's unconditionally convergent but not absolutely convergent?
The two properties are equivalent in R and in finitely-dimensional spaces.

Try \(\displaystyle X=\ell_2\) and the series \(\displaystyle x_n=\frac 1{\sqrt{n}}e_n\), where \(\displaystyle e_n\) denotes the sequence which has one on n'th place and all other terms are zeros.

If I remember correctly, this is precisely the most basic example given in the book
Kadets, Kadets: Series in Banach spaces: conditional and unconditional convergence
which is devoted basically to this topic.

However, many texts on functional analysis mention the relationship between various modes of convergence in Banach spaces.
I remember I have seen this in Wojtaszczyk's Banach spaces for analysts (with proofs) and in Megginson's Intorduction to Banach spaces theory (as an exercise - if I remember correctly).