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The Weierstrass M-Test states that if you have a sequence of functions $(f_k)_{k\epsilon\mathbb N}$ where $f_k:A\mapsto\mathbb R$, and suppose that $\forall k \epsilon \mathbb N \exists M_k >0 $ such that

$$|f_k(x)|\leq M_k \forall x \epsilon A$$ and $$\sum_{k=1}^{\infty}M_k <\infty$$

Then the series $\sum_{k}f_k$ converges uniformly on A.

However, the Weierstrass M-test is not applicable to series that converge uniformly but not absolutely.

Therefore, when attempting to prove a series converges uniformly using the M test, do I need to show first of all that the series converges absolutely?

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  • $\begingroup$ If you've shown $|f_k|\le M_k$ and $\sum M_k<\infty$, then you've already shown it: the series converges absolutely everywhere by the comparison test. $\endgroup$ – zhw. Dec 19 '17 at 18:53
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Not really. The conclusion of the usual statement of the M test (cf Wikipedia) is that the series converges uniformly and absolutely. So obviously if it does not converge absolutely you will never be able to use the M test, but that will be reflected in the premises of the M test not holding.

In particular, if $f_n(x)$ did not converge absolutely for some $x,$ and the premises of the m test held we would have $$\infty=\sum_n |f_n(x)| \leq \sum_n M_n < \infty,$$ a clear contradiction.

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  • $\begingroup$ So if I have a series of functions and I have to decide whether it converges uniformly or not, I could first try using the M-test. If the M-test conditions are not satisfied, this does not necessarily mean the series does not uniformly converge. However I don't need to check whether the series converges absolutely as if it doesn't, it would fail the conditions for the M-test and I would have to use another test to decide whether it is uniformly convergent. Is this correct? $\endgroup$ – Thomas Smith Dec 19 '17 at 18:06
  • $\begingroup$ Yes that's right. For instance I saw this question the other day about proving uniform convergence for a series that didn't converge absolutely math.stackexchange.com/questions/2567371/… . They needed an different approach than the M test. $\endgroup$ – spaceisdarkgreen Dec 19 '17 at 18:19

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