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For what $x \in \mathbb{R}$ or interval $I \subset \mathbb{R}$ is the following series uniformly convergent?

$$\sum_1^{\infty}\frac{x^n}{1+x^{2n}}$$

My hunch has been to try and find some way to express this in the standard power series form and find the radius of convergence. Tried using the definition of uniform convergence of sequences of functions as well by considering the partial sums and attempting a bounding but didn't work either.

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I will prove that the series is well defined and converges uniformly on compacts on the open set $\mathbb R\setminus \{-1,1\}$.

Let $$f_n (x) = \frac{x^n}{1+x^{2n}}.$$ Notice that: $$\frac{f_{n+1}(x)}{f_n(x)} = \frac{x(x^{2n}+1)}{x^{2n+2}+1}.$$

If $|x|>1$ then $$\lim_{n\to \infty} \left|\frac{f_{n+1}(x)}{f_n(x)}\right| = 0,$$ and therefore $\sum_{n=1}^{\infty} f_n(x)$ converges.

If $|x|<1$ then $$\lim_{n\to \infty} \left|\frac{f_{n+1}(x)}{f_n(x)}\right| = |x| < 1,$$ and therefore the above series converges.

So, the series is well defined on $I=\mathbb R \setminus \{-1,1\}$. Notice that the series does not converge for $|x|=1$ because $|f_n(x)| = 1 \not\to 0$.

Let $g_n(x) = \sum_{k=1}^n f_k(x)$.

If $k<l$ we have $$|g_k(x) - g_l(x)| \leq \sum_{j=k+1}^l\frac{|x|^j}{1+x^{2j}}.$$

If $K$ is a compact subset of $(-1,1)$ then there is $0<r<1$ such that $K\subset [-r,r]$. Then we have $$\sup_{|x|\leq r}|g_k(x)-g_l(x)|\leq \sum_{j=k+1}^l \frac{r^j}{1+r^{2j}}\leq \sum_{j=k+1}^l r^j \to 0$$ when $k,l \to \infty.$

So, $g_n$ converges uniformly on each compact contained in $(-1,1)$.

If $K$ is a compact set contained in $\mathbb R \setminus [-1,1]$ then there is $r>1$ such that $|x|>r$ for each $x\in K$. For $|x|>r$ we have

$$|g_k(x)-g_l(x)|\leq \sum_{j=k+1}^l \frac{|x|^j}{1+x^{2j}}\leq \sum_{j=k+1}^l \frac{1}{|x|^j}\leq\sum_{j=k+1}^l \frac{1}{r^j} \to 0$$ when $k,l \to \infty$. In the above calculation I am using that $1+x^{2j}\geq x^{2j}$. Therefore we obtain $$\sup_{r\leq |x|\leq R}|g_k(x)-g_l(x)|\leq \sum_{j=k+1}^l \frac{1}{r^j} \to 0$$ when $k,l \to \infty$.

Therefore, $g_n$ converges uniformly on each compact contained of $\mathbb R \setminus \{-1,1\}.$

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Hint:

Consider separately the cases $x<-1$, $-1<x<1$, $x>1$. Weierstrass M-test suffices to show uniformly convergence at these intervals.

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