I have solved the following exercise and I would like to know if I have made any mistakes:
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 continuos function on this set.
My solution:
$|x|<1\Rightarrow 0<x^{2n}<1\Rightarrow 1<1+x^{2n}<2\Rightarrow \frac{1}{2}<\frac{1}{1+x^{2n}}<2\Rightarrow 0<\frac{x^{2n}}{2}<\frac{x^{2n}}{1+x^{2n}}<x^{2n}$ so by Comparison Test $\sum_{n=0}^{\infty}\frac{x^{2n}}{1+x^{2n}}$ converges for all $x\in (-1,1)$ (since $\sum_{n=0}^{\infty} x^{2n}$ converges in that interval to $\frac{1}{1-x^2}$). If $|x|=1$ then $\lim_{n\to\infty}\frac{x^{2n}}{1+x^{2n}}=\frac{1}{2}\neq 0$ so the series does not converge and if $|x|>1$ then $\lim_{n\to\infty}\frac{x^{2n}}{1+x^{2n}}=\lim_{n\to\infty}(1-\frac{1}{1+x^{2n}})=1\neq 0$ so the series does not converge here too; the series thus converges only if $|x|<1$.
If $x\in [-M,M], |M|<1$ then $\frac{x^{2n}}{1+x^{2n}}\leq x^{2n}\leq M^{2n}$ and since $\sum_{n=0}^{\infty} M^{2n}=\frac{1}{1-M^2}$ by Weierstrass M-Test the series converges uniformly on each interval $[-M,M], |M|<1$ and so by Term-by-Term Continuity Theorem $g(x)$ is continuous on every such interval.
It is correct, except that you wrote nothing about what happens when $x=-1$. In fact, in that case we have $(\forall n\in\Bbb N):\frac{x^{2n}}{1+x^{2n}}=\frac12$, and therefore the series diverges then too. Actually, that argument also works when $x=1$ too.