Prove uniform convergence of a series Define $g(x)$ = $\sum_{n=0}^\infty \frac{x^{2n}}{1+x^{2n}}$. Find the values of x where the series converges and show that we get a continuous function on this set.
So I know that if I can prove uniform convergence, then that implies uniform convergence. Is there a way to use the Weierstrass M-test here? I also can't figure out what to do to find the radius of convergence.
 A: We have $g(x) = \sum_{n=0}^\infty \frac{x^{2n}}{1+x^{2n}} = \sum_{n=0}^\infty \frac{1}{1+x^{-2n}} < \sum_{n=0}^\infty \frac{1}{x^{-2n}} = \sum_{n=0}^\infty x^{2n} = \sum_{n=0}^\infty (x^2)^n$.
This is a geometric series, so it will converge when $|x^2| < 1 \rightarrow x \in (-1,1)$.
For $|x| = 1$, we have $g(x) = \sum_{n=0}^\infty \frac{1}{2}$, so the series is divergent.
For $|x| > 1$, we have $\lim_{n\rightarrow \infty} \frac{x^{2n}}{1+x^{2n}}=1\neq0$, so the series is divergent. 
A: Another way to see the series doesn't converge uniformly on $(-1,1):$ If $\sum_n f_n$ converges uniformly on a set $E,$ then $f_n \to 0$ uniformly on $E,$ which is the same as saying $\sup_E |f_n| \to 0.$ But in our case,
$$\sup_{(-1,1)} \frac{x^{2n}}{1+x^{2n}} \ge \frac{1}{2},$$
so uniform convergence fails.
A: To address the question on uniform convergence, let $\epsilon>0$ be given,  Then, for any $a<1$, note that we have for $x\in [-a,a]$
$$\begin{align}
f_n(x)&=\sum_{k=n}^\infty \frac{x^{2k}}{1+x^{2k}}\\\\
&\le \sum_{k=n}^\infty a^{2k}\\\\
&=\frac{a^{2n}}{1-a^2}\\\\
&<\epsilon
\end{align}$$
whenever $n>\frac{\log((1-a^2)\epsilon)}{2\log(a)}$.  So, the convergence is uniform for $x\in [-a,a]$ for any $a<1$.
However, the convergence is not uniform for $x\in (-1,1)$.  Take $\epsilon=1/32$.  Then, note that for $x=1-\frac1n$ we have for $n\ge 2$
$$\begin{align}
f_n(x)&=\sum_{k=n}^\infty \frac{x^{2k}}{1+x^{2k}}\\\\
&\ge \frac12 \sum_{k=n}^\infty x^{2k}\\\\
&=\frac12 \frac{x^{2n}}{1-x^2}\\\\
&\ge\frac12 \left(1-\frac{1}{n}\right)^{2n}\\\\
&\ge \frac{1}{32}
\end{align}$$
