Talk:Weierstrass M-test

I'm reading Complex Analysis by Freitag and Busam, and their version of the Weiestrass M-test say that every normally convergent series converges absolutely and locally uniformly. Their definition of normal convergence also differs with the definition found on Wikipedia. They say that normal convergence of fn: D -> C means that for every a in D, there exists a neighborhood of a, U, such that there exists a sequence (of real numbers) Mn such that for every z in intersection(D, U), abs(fn(z)) < Mn and SUM(Mn) converges.

One thing I'm wondering is, what do they mean by "absolutely convergent". One clue is that they say a consequence of the Weiestrass M-test is that normally convergent series can be arbitrarily reordered without disturbing the limit of the series... Danielx 06:11, 7 March 2007 (UTC)[reply]

See absolute convergence. Oleg Alexandrov (talk) 15:34, 7 March 2007 (UTC)[reply]

This article should probably be merged with Normal_convergence, as the hypothesis of the M-test theorem and the concept of normal convergence coincide. 85.49.191.6 (talk) 01:35, 24 April 2014 (UTC)[reply]

Centralized discussion on proofs

See WT:MATH#Proofs, revisited — Arthur Rubin (talk) 17:58, 29 September 2015 (UTC)[reply]

Need to show "uniformly Cauchy" implies "uniformly convergent"

We need to prove the "obvious" result that if a sequence is "uniformly" Cauchy, then it uniformly converges. It pointwise converges by completeness, and then uniformly converges as (approximately)

If $\lim _{n}f_{n}(x)=f(x)$ , then $\lim _{m}|f_{n}(x)-f_{m}(x)|=|f_{n}(x)-f(x)|$ , so if $\forall \varepsilon >0\exists N\forall x\forall m,n>N|f_{n}(x)-f_{m}(x)|<\varepsilon$ , then, using the same N, $\forall \varepsilon >0\exists N\forall x\forall n>N|f_{n}(x)-f(x)|\leq \varepsilon$ . But we must say that somewhere, for the proof to be complete. — Arthur Rubin (talk) 15:33, 30 September 2015 (UTC)[reply]