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Re: Exchanging the order of summation
Posted:
Nov 10, 2011 6:52 AM
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On Nov 9, 11:08 am, "G. A. Edgar" <ed...@math.ohio-state.edu.invalid>
wrote:
> In article
> <d76d43c30a634c5cb29e6ab44da4d...@CITESHT4.ad.uillinois.edu>, John
>
> Washburn <M...@WashburnResearch.org> wrote:
> > Are there conditions other than uniform convergence or absolut
> > convergence, which permit the order of summation to interchanged?
>
> A simple example to consider is this one: f(n,n)=1, f(n,n+1)=-1,
> for all n, and everything else 0. Interchanged sums are not equal.
>
>
>
> >
> >
> > I have a double summation over n = 1 to \infty and q= 1 to \infty of
> > the summand f(n,q). The limit processess are q first, then n, but i
> > would like to evaluate n first then q. If it matters f(n,q) is finite
> > and real for positive integers, n and q.
> >
> > I have sum with a definite when there is a single limit process
> > involved. Namely, I have two non-decreasing functions g(Q) and h(Q)
> > and a well define limit as Q increases without bound:
> >
> > limit_{Q \to \infty} sum_{n=1}^{g(Q)} sum_{q=1}^{g(Q)} = K.
> >
> > I seems to me I am very close to the Fubini-Tonelli theorem and that
> > if the double summation with a single limit process has a finite limit
> > the iterated sum has the same finite limit regardless of the order of
> > summation.
> >
> > Or is the proper conclusion that if a finite, limit exists, then all
> > three limits are the same. No guarantee that a finite limit exist,
> > jsut that if it does all three limit processes lead to the same value.
> >
> > So my question in another form is this:
> > Is the existence of a finite value of the double sum using a single
> > limit process (functions of Q), sufficient to permit the interchanging
> > the order of the limit processes; q tends to infinity and n tending to
> > infinity?
> >
> > Thanks for any time you might give to this question.
> > John Washburn
I should have wrote:
limit_{Q \to \infty} sum_{n=1}^{g(Q)} sum_{q=1}^{g(Q)} = K.
as:
limit_{Q \to \infty} sum_{n=1}^{g(Q)} sum_{q=1}^{h(Q)} f(n,q) = K.
Sorry for the typing error.
It looks like uniform convergence though is required, but the Moore-
smith theorem may be applicable as the sequence as n tends to infinity
coverges uniformly and point-wise as q tends to infinity.
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