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Re: Exchanging the order of summation
Posted:
Nov 10, 2011 6:52 AM


On Nov 9, 11:08 am, "G. A. Edgar" <ed...@math.ohiostate.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 nondecreasing 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 FubiniTonelli 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 pointwise as q tends to infinity.



