1
$\begingroup$

I have come across an exercise which asks to prove that the series of functions$$\sum\frac{x^n}{1+x^n}$$ is convergent for $x\in [0,1)$. It also asks us to prove that the series converges uniformly for each $a: 0<a<1$. I think I know how to answer the second question (proving the uniform convergence of the series): $$\sum\frac{x^n}{1+x^n}\leq\sum x^n$$ We can then use the Weierstrass M-test and let $\sum M_k$= $\sum x^n$, which then converges to $\frac{1}{1-x}$ for $\vert x\vert<1$. Hence $\sum\frac{x^n}{1+x^n}$ is uniformly convergent. However, I am not sure how to answer the first question. I think that I could just say that since the series is uniformly convergent, it must converge. Or I could use the same inequality as above and use the comparison test to show that it converges. However, I am not sure. What am I supposed to do? Is it asking whether the series of functions is pointwise convergent? If not what is the difference between proving that a series of functions converges pointwise vs. proving that a series of functions converges? Am I missing something? Also, is my approach in proving uniform convergence correct?

$\endgroup$
  • $\begingroup$ What series of functions? Where is $a$ used? Did you mean to have $x^n$ instead of $x_n$? $\endgroup$ – copper.hat Nov 17 '13 at 8:29
  • $\begingroup$ Yes, sorry. It should be a superscript. $\endgroup$ – CoffeeIsLife Nov 17 '13 at 8:37
  • $\begingroup$ Also, $a$ should be the value of $x$ $\endgroup$ – CoffeeIsLife Nov 17 '13 at 8:48
  • $\begingroup$ Did you leave something out of the problem statement? Were you asked to prove uniform convergence on the set [0,a] for each $a$? Is that what you proved, or did you prove more, that the series converges uniformly on the set $[0,1)$? $\endgroup$ – bof Nov 17 '13 at 9:13
  • 1
    $\begingroup$ @No, it’s not the case. But there is no difficulty indeed, in some way it is simpler to answer the second question before the first. What happens is this : the series converges uniformly on every $[0,a]$, so it converges simply on every $[0,a]$, so it must converge simply on the union of all those $[0,a]$'s, which is $[0,1)$. But it is not true that the series converges uniformly on $[0,1)$. If it did, it would converge at $x=1$ also. $\endgroup$ – Ewan Delanoy Nov 17 '13 at 12:04
1
$\begingroup$

We can then use the Weierstrass $M$-test and let $\sum M_k= \sum x^n$

No. The whole point of $M$-test is that $M$-numbers do not involve $x$. What should be done instead: on the interval $[0,a]$ we have $$\frac{x^n}{1+x^n}\le x^n \le a^n$$ Let $M_n=a^n$ and apply the test.

I think that I could just say that since the series is uniformly convergent, it must converge.

Yes. More precisely: every point $x\in [0,1)$ is contained in some interval $[0,a]$ with $a<1$ (take any $a$ with $x<a<1$), and we already know that the series converges, even uniformly, on $[0,a]$. Thus it converges at $x$; and this is all we need to show to demonstrate pointwise convergence on $[0,1)$.


Apparently, you were not asked to show that uniform convergence fails on $[0,1)$. For the same of completeness, here is a sketch: for every $n$ there is $x\in [0,1)$ such that $x^n=1/2$. Hence $\frac{x^n}{1+x^n}=1/3$. So, the terms of the series do not tend to $0$ uniformly, thus failing a necessary condition for uniform convergence.

$\endgroup$
  • $\begingroup$ Nice :). We can also prove the uniform convergence by proving the normal convergence on $[0,a]$ $\endgroup$ – Nour May 17 '17 at 14:05

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.