1
$\begingroup$

I think this has to do with Dini's theorem, because I notice that $f_n(x)= \frac{x^n}{1+x^{2n}}$ is continuous and monotonically converges because of: $$\lim_{n \rightarrow \infty} f_n(x) = 0$$ and $$f_n(x) \geq f_{n+1}(x)$$ But then question asks for the convergence of a series, so I am not sure how to go about this.

I also tried the Weierstrass M-test with some $p$-series and Cauchy criterion for uniform convergence, but the "every compact subset" just does not seem to fit.

$\endgroup$
  • 1
    $\begingroup$ How does this series converge for $x=1$? $\endgroup$ – Mark Viola Nov 14 '15 at 5:22
  • $\begingroup$ Possible Duplicate of THIS. $\endgroup$ – Mark Viola Nov 14 '15 at 5:28
  • $\begingroup$ It should be $[0,\infty)\backslash\{1\}$. The other question has some similarities, but my question is specifically about the every compact subset part (which leads to uniform convergence). $\endgroup$ – Paichu Nov 14 '15 at 5:36
  • $\begingroup$ Take any number $x_0<1$. Then on any closed set $[0,x_0]$, it is easy to show UC. Then, take any numbers $1<x_1<x_2<\infty$, it is easy to show UC on $[x_1,x_2]$. Does this help? $\endgroup$ – Mark Viola Nov 14 '15 at 5:38
  • $\begingroup$ Yes, however, I am a bit not sure if this includes every compact subsets. For example, if I take the cantor set between $[x_1, x_2]$, would I still be able to show UC? $\endgroup$ – Paichu Nov 14 '15 at 5:46
2
$\begingroup$

Take $x_0<1$. Then on any compact set $[0,x_0]$, we have

$$\frac{x^n}{1+x^{2n}}\le x_0^n$$

Since $\sum_{n=0}^\infty x_0^n=\frac{1}{1-x_0}<\infty$, then by the Wierestrass M Test, the series $ \sum_{n=0}^\infty \frac{x^n}{1+x^{2n}}$ converges uniformly on $[0,x_0]$ for all $x_0<1$.

Now, take any two numbers $1<x_1<x_2<\infty$. On any compact set $[x_1,x_2]$, we have

$$\frac{x^n}{1+x^{2n}}\le\frac{1}{x_n}\le \frac{1}{x_1^n}$$

$\sum_{n=0}^\infty \frac{1}{x_1^n}=\frac{x_1}{x_1-1}<\infty$, then by the Wierestrass M Test, the series $ \sum_{n=0}^\infty \frac{x^n}{1+x^n}$ converges uniformly .

$\endgroup$

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.