Uniform convergence of series of functions

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SUMMARY

The series \(\sum \frac{x^n}{1 + x^n}\) is not uniformly convergent on the interval [0,1). The Weierstrass M-Test confirms uniform convergence on any subinterval [0,a] for \(0 < a < 1\). However, applying the Cauchy criterion reveals that the supremum \(\sup \{|\frac{x^n}{1 + x^n}| : x \in [0,1)\}\) equals \(\frac{1}{2}\), which does not approach zero as \(n\) approaches infinity, thus proving the series fails to be uniformly convergent on [0,1).

PREREQUISITES
  • Understanding of uniform convergence in the context of series of functions
  • Familiarity with the Weierstrass M-Test for uniform convergence
  • Knowledge of the Cauchy criterion for uniform convergence
  • Basic calculus, including differentiation and finding maxima/minima of functions
NEXT STEPS
  • Study the Weierstrass M-Test in detail to understand its applications
  • Learn about the Cauchy criterion for uniform convergence and its implications
  • Explore examples of series of functions that are uniformly convergent versus those that are not
  • Investigate the implications of uniform convergence on the interchange of limits and integrals
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Mathematics students, particularly those studying real analysis, and educators looking to deepen their understanding of convergence concepts in series of functions.

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Homework Statement


Is the series \sum \frac{x^n}{1 + x^n} uniformly convergent on [0,1) ?

Homework Equations


The Attempt at a Solution


I'm guessing that it is NOT uniformly convergent but I don't know how to show it. But I would note that it is quite easy to show that the series is uniformly convergent on [0,a] for \forall a, 0 &lt; a &lt; 1 (i.e. straightforward application of the Weierstrass M-Test).

I'm aware of the following fact is useful to check whether a series of functions is "not" uniformly convergent. We know that if \sum g_k is uniformly convergent on S \subseteq \mathbb{R}, then \lim_{n \to \infty} \sup \{|g_k(x)| : x \in S\} = 0.

Hence motivated by this fact, a good trick would be to use derivatives to find the maximum. But in this case, it doesn't work so well. For convienence, set g_n(x) = \frac{x^n}{1 + x^n} and in this case we can drop the absolute value since x \in [0,1). And if my differentation is correct, we should have g_n&#039;(x) = \frac{nx^{n-1}}{(1 + x^n)^2} and set this to zero to find the maximum / minimum and we see that g_n&#039;(x) = 0 implies x = 0 --- and I'm very certain that x = 0, so that g_n(x) = 0, is not the maximum of g_n, and in fact it is clear that it is the minimum. Thus, the above attempt to find the supremum doesn't work so well.

I also thought about using the Cauchy criterion for uniform convergence, but there does not seem to be any "convenient" indicies for me to choose to show that it fails the criterion.Any help is greatly appreciated!
 
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Given k, what is the set:

\sup \{|g_k(x)| : x \in [0,\, 1)\}

Find out what that set is, then it's easy to compute its sup, and then computing the lim of those sups is going to be easy, and your proof will be done.
 
AKG said:
Given k, what is the set:

\sup \{|g_k(x)| : x \in [0,\, 1)\}

Find out what that set is, then it's easy to compute its sup, and then computing the lim of those sups is going to be easy, and your proof will be done.

Thanks for the reply. I thought a little bit more on the problem yesterday and I think I made it way too complicated in my previous attempt.

It's simply as follows. The number is \sup \{|\frac{x^n}{1 + x^n}| : x \in [0,1)\} = \frac{1}{2} and this is seen by setting x = 1. And of course, if that is equal to 1/2, then its limit is still 1/2 and thus, this series cannot be uniformly convergent.
 

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