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 $x$.)


  • 2
    $\begingroup$ "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

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.