Is there a series of functions which converges uniformly and absolutely but does not pass the weirstrass m-test?
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Related question http://math.stackexchange.com/questions/26273/converse-of-the-weierstrass-m-test?rq=1 – Winther Nov 11 '15 at 00:37
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Good question in the sense that we are trained, even in a sleepy or inebriated state to say "What about the converse?" (Not necessarily socially appropriate in all situations but mathematicians are usually oblivious anyway.)
Take the sequence of bounded functions $f_n(x)=1/n$ for $n-1\leq x< n $ and zero otherwise ($n=1,2,3, \dots)$. On the interval $[0,\infty)$ the series $\sum_{n=1}^\infty f_n(x)$ converges absolutely and uniformly. But $$\sum_{n=1}^\infty \|f_n\|_\infty = \sum_{n=1}^\infty 1/n =\infty. $$
[Note added: I see that this does have an answer already somewhere else. Maybe all questions have been asked before and there is nothing to be asked here except research questions!]
B. S. Thomson
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For example, consider $$ f_1(x)=x\\ f_n(x)= \frac 1{2^n} \quad n \geq 2 $$ with domain $\Bbb R$.
Ben Grossmann
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1Not an answer within the spirit of the question however. Here you have the series $$f_1(x) + \sum_{n=1}^\infty f_{n+1}(x)$$ and the latter part does converge uniformly by the M-test. – B. S. Thomson Nov 11 '15 at 00:35
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No, I think Om is right. This series converges uniformly, and absolutely, on $\mathbb R$ but the first summand is unbounded, hence fails Weirstrass M. – zhw. Nov 11 '15 at 02:13
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1That's not the point. Here is a version of the M-test that this example won't work on: A series $\sum_{n=1}^\infty f_n(x)$ converges absolutely and uniformly if there is an integer $N$ so that $\sum_{n=N}^\infty|f_n|_\infty<\infty$. Who cares what the first 700 terms of the series are doing? Would you really give up in a situation where the first few terms were unbounded and say "Alas the M-test has failed me!" – B. S. Thomson Nov 11 '15 at 03:56
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