For which intervals on $\Bbb R$ is the following series uniformly convergent?
$$\sum_{n=1}^\infty\frac{x^n}{1+x^{2n}}$$
This is what I thought:
Let $a\in[0,1)$. Then on intervals $\left[-a,a\right]$:
$$\left|\frac{x^n}{1+x^{2n}}\right|\leq a^n$$ And the series $\sum a^n$ converges. It is uniformly convergent by the M test. Is this correct ? Are there more intervals ?
Yes, this is correct.
Now note the series diverges for $x=\pm 1$.
And for all $|x|> a>1$, you have $$ 0\leq \frac{x^n}{1+x^{2n}}\leq \frac{x^n}{x^{2n}}= \left(\frac{1}{x}\right)^n\leq \left(\frac{1}{a}\right)^n. $$ So you have also uniform convergence by the $M$-test on $(-\infty,-a]\cup[a,+\infty)$.
You can't do better than these intervals, ie you don't have uniform convergence on $(-1,1)$, $(-\infty,-1)$ or $(1,+\infty)$.
Otherwise, by the uniform Cauchy criterion, you would find for instance that $$ \frac{1}{2}=\lim_{n\rightarrow+\infty} \sup_{x\in(-1,1)}\frac{x^n}{1+x^{2n}}=0. $$ Contradiction. It works the same on the other two intervals.
