On Tuesday, April 9, 2013 5:14:25 PM UTC+1, dull...@sprynet.com wrote: > On Mon, 8 Apr 2013 04:33:13 -0700 (PDT), firstname.lastname@example.org wrote: > > > > >I have difficulty following the proof of theorem 8.3 in the 3rd edition of Principles of Mathematical Analysis by Rudin. The line I'm struggling with is: > > >lim n -> infinity [ sum(i = 1 to infinity) [sum j = 1 to n a_ij ] ] = > > >= lim n -> infinity [sum j = 1 to n [sum i = 1 to infinity a_ij ] ] > > > > > >This doesn't seem immediate, and the way I prove this is by theorem 3.55. > > > > From your comments below I gather that 3.55 has something to do with > > rearranging absolutely convergent series? That has nothing at all to > > do with this, that I can see. The equality you ask about is entirely > > triivial. > > > > I don't have the book with me, so I can't say what the section numbers > > are. But surely one of the very first things he proves about infinite > > sums is this: > > > > Lemma. If sum_1^infinity a_j and sum_1^infinity b_j both > > converge then sum_1^infinity (a_j + b_j) converges and > > equals sum a_j + sum b_j. > > > > The lemma is entiirely trivial, has nothing to do with absolute > > convergence or rearrangements, just the definitions (hint: > > epsilon = epsilon/2 + epsilon/2). > > > > And it's immediate from the lemma by induction on n that > > > > sum(i = 1 to infinity) [sum j = 1 to n a_ij ] = > > = sum j = 1 to n [sum i = 1 to infinity a_ij ] . > > > > Hence lim_n of both sides is the same. > > > > >However, the whole statement of 8.3 seems immediate from 3.55 anyway, since you just need to arrange the countable number of terms a_ij into a single sequence and appeal to the absolute convergence. > > > > > >So Rudin's exposition of theorem 8.3 doesn't help me. I can only follow it by using theorem 3.55 and I understand that his intention is to avoid theorem 3.55. > > > > > >What am I missing? > > > > > >Thank You, > > > > > >Paul Epstein
I agree with you, but I resolved the issue myself in my 2nd posting.