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Nested series

  1. Feb 23, 2007 #1


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    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]
    Last edited: Feb 23, 2007
  2. jcsd
  3. Feb 23, 2007 #2
    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.
  4. Feb 24, 2007 #3

    Gib Z

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    Yes there is a question here, see statement 4.
  5. Feb 24, 2007 #4
    answer to statement !?

    I reads like an application of the Fubini -Tonelli Thm. (and no question mark).
  6. Feb 24, 2007 #5

    Gib Z

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    "Can we conclude that...."...if your picky stick a question mark on the end.
  7. Feb 24, 2007 #6
  8. Feb 24, 2007 #7


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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!)
  9. Feb 24, 2007 #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...

    each horizontal sum is zero. The vertical sums are alternately 1,-1,1,-1,... which doesn't converge, never mind absolutely.
  10. Feb 24, 2007 #9


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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?
    Last edited by a moderator: May 2, 2017
  11. Feb 24, 2007 #10
    Now the question makes sense, the statement 3) was not clear.
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