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I have a simple question about the Weierstrass M-test in the context of real analysis on the line. The Weierstrass M-test can be stated as:

Let $\{f_n\}$ be a sequence of real functions defined in a subset $D$ of the real line. Suppose that $\forall n\in \mathbb{N}$ and $\forall x\in D$ , $| f_n(x) | \leq A_n$ . If $\sum A_n$ converges then $\sum f_n $ converges uniformly on $D$.

I would like to know if the hypotheses $\forall n\in \mathbb{N}$ and $\forall x\in D$ , $| f_n(x) | \leq A_n$ is necessary. Is possible to suppose only that $\forall x\in D$ , $| f_n(x) | \leq A_n$ for $n$ large enough? (n is large enough but it is still independent of $x$.)

Thanks!

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    $\begingroup$ "For $n$ large enough" may be ambiguous as it doesn't tell whether the "enough" is uniform w.r.t. $x$ or depends pointwise on $x$, which is all the difference here. $\endgroup$ – Vim Feb 21 '16 at 13:24
  • $\begingroup$ Yes, you can assume so, because uniform convergence doesn't depend break upon changing any finite number of functions in the sum. $\endgroup$ – Wojowu Feb 21 '16 at 13:25
  • $\begingroup$ Dear Vim n is large enough but independent of $x$. I edited the original question. Thanks Wojowu for your answer. $\endgroup$ – bebop Feb 21 '16 at 13:31
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It depends exactly "how badly" you contravene the hypotheses:

  • If any of the functions $f_{n}$ fails to be defined at points of $D$, then convergence of the series is out.

  • If your functions are bounded but happen to be larger than your given sequence $(A_{n})$, there's no problem with assuming only "for sufficiently large $n$": Changing finitely many terms of an infinite numerical series has no effect on convergence/divergence.

  • If your functions are defined everywhere but only finitely many are unbounded (say $f_{n}$ is bounded for $n \geq N$), you're still OK: Split the series into the first $N$ terms (whose sum is an algebraic matter), and the remaining terms, to which the $M$-test applies.

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