# Show $\sum_{n=0}^{\infty} \frac{x^n}{1+x^{2n}}$ converges uniformly on every compact subset of $\,\left[0,\,\infty\right)\setminus\left\{1\right\}\,$.

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.

• How does this series converge for $x=1$? – Mark Viola Nov 14 '15 at 5:22
• Possible Duplicate of THIS. – Mark Viola Nov 14 '15 at 5:28
• 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). – Paichu Nov 14 '15 at 5:36
• 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? – Mark Viola Nov 14 '15 at 5:38
• 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? – Paichu Nov 14 '15 at 5:46

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 .