Nested series

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AKG
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Main Question or Discussion Point

Let [itex]z : \mathbb{N}^2 \to \mathbb{C}[/itex]. Suppose that:

1) for all natural n, [itex]\sum _{j \in \mathbb{N}}z(n,j)[/itex] converges absolutely.

2) for all natural j, [itex]\sum _{n \in \mathbb{N}}z(n,j)[/itex] converges absolutely.

3) [itex]\sum _{n \in \mathbb{N}}\left (\sum _{j \in \mathbb{N}} z(n,j)\right )[/itex] converges absolutely.

Can we conclude that

4) [itex]\sum _{j \in \mathbb{N}}\left (\sum _{n \in \mathbb{N}} z(n,j)\right )[/itex] converges absolutely as well, with

[tex]\sum _{j \in \mathbb{N}}\left (\sum _{n \in \mathbb{N}} z(n,j)\right ) = \sum _{n \in \mathbb{N}}\left (\sum _{j \in \mathbb{N}} z(n,j)\right )[/tex]
 
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  • #2
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Is there a question here? If you want to prove it, see Folland's "Real Analysis: Modern Techniques and Their Applications. For errata, see Folland's website.
 
  • #3
Gib Z
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Yes there is a question here, see statement 4.
 
  • #4
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answer to statement !?

I reads like an application of the Fubini -Tonelli Thm. (and no question mark).
 
  • #5
Gib Z
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"Can we conclude that...."...if your picky stick a question mark on the end.
 
  • #6
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Let [itex]z : \mathbb{N}^2 \to \mathbb{C}[/itex]. Suppose that:

Can we conclude that

4) [itex]\sum _{j \in \mathbb{N}}\left (\sum _{n \in \mathbb{N}} z(n,j)\right )[/itex] converges absolutely as well, with

[tex]\sum _{j \in \mathbb{N}}\left (\sum _{n \in \mathbb{N}} z(n,j)\right ) = \sum _{n \in \mathbb{N}}\left (\sum _{j \in \mathbb{N}} z(n,j)\right )[/tex]
?
:surprised
 
  • #7
Hurkyl
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I claim there exists a z such that:

[tex]\forall n: \sum_j z(n, j) = 0 [/tex]

[tex]\forall j: \sum_n z(n, j) = 1 [/tex]

are all absolutely convergent sums. And clearly a sum of zeroes is absolutely convergent...


If you're going to come up with a counter-example, look for a simple one! I didn't come up with this one until I stopped trying to make clever counter-examples.




(In fact, we can arrange it so that each sum has a finite number of nonzeroes!)
 
  • #8
matt grime
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The way to think about counter examples for these is to think about the lattice of NxN, and summing rows, then columns etc. It is usually quite easy to come up with counter examples. My counter example was this:

z(1,j)= 1-1+0+0+0...
z(2,j)=0+0+1-1+0+0+.....


each horizontal sum is zero. The vertical sums are alternately 1,-1,1,-1,... which doesn't converge, never mind absolutely.
 
  • #9
AKG
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Thanks guys! I had a problem where part of it had a z(n,j) satisfying 1-3 and I needed it to satisfy 4. Fortunately, I had other hypotheses to work with and I ended up using those to get 4. But it was good seeing your counterexamples. In general, what conditions do we need to get 4? PlanetMath gives a sufficient condition http://planetmath.org/encyclopedia/DoubleSeries.html [Broken], is there a concise way to express a necessary and sufficient condition?
 
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  • #10
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Now the question makes sense, the statement 3) was not clear.
 

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