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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 ?

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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.

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