I am self-learning real-analysis from the text, Understanding Analysis by Stephen Abbott. I'd like for someone to verify, if my deduction below is rigorous and correct.
[Abbott 6.4.4] Define:
$$g( x) =\sum _{n=0}^{\infty }\frac{x^{2n}}{1+x^{2n}}$$
Find the values of $\displaystyle x$ where the series converges and show that we get a continuous function on this set.
Proof.
Let $\displaystyle x$ belong to any compact interval $\displaystyle [ a,b]$ such that $\displaystyle -1< a< b< 1$. Let $\displaystyle c=\max\{|a|,|b|\}$.
We have:
$$g_{n}( x) =\frac{x^{2n}}{1+x^{2n}} \leq x^{2n} \leq c^{2n}$$
Define $M_{n} =c^{2n}$. Then,
$$\sum _{n=1}^{\infty} M_n =\frac{1}{1-c^{2}}$$
By the Weierstrass $\displaystyle M$-test, the infinite series $\displaystyle \sum _{n=1}^{\infty }\frac{x^{2n}}{1+x^{2n}}$ converges uniformly on $\displaystyle [a,b]$.
Since $\sum_{n=1}^{\infty} g_{n}$ converges uniformly on $[a,b]$ to $\displaystyle g$, and each $\displaystyle g_{n}$ is continuous on $\displaystyle [ a,b]$, by the term-by-term continuity theorem, $\displaystyle g$ is continuous on $\displaystyle [a,b]$.
Also, $\sum_{n=1}^{\infty}x^{2n}$ converges pointwise on $(-1,1)$. So, by the comparison test, $\sum_{n=1}^{\infty}\frac{x^{2n}}{1+x^{2n}}$ converges pointwise on $(-1,1)$.
The radius of the convergence might be greater than 1.
proposing another way that you can do it:
Definition: $$f(x) = \sum_{n=1} ^{\infty} \frac{x^{2n}}{1 + x^{2n}}$$
#case 1: $\lvert x \rvert > 1 \to \displaystyle \lim_{n \to \infty} x^{2n}=\infty$: $$\lim_{n\to\infty} \frac{x^{2n}}{1 + x^{2n}} = 1$$
#case 2: $\lvert x \rvert =1$ $$\lim_{n \to\infty} \frac{x^{2n}}{1 + x^{2n}} = \frac{1}{2}$$
Then, for $\lvert x \rvert \ge 1$, $\sum_{n=1} ^{\infty} \frac{x^{2n}}{1 + x^{2n}}$ does not converge, and for $\lvert x \rvert < 1$: $$\sum_{n=1} ^{\infty} \frac{x^{2n}}{1 + x^{2n}} \leq \sum_{n=1} ^{\infty} x^{2n}$$
And since $\sum_{n=1} ^{\infty} (x^2)^n$ is a geometric sequence with $x^2 < 1$, it converges, thous, $f(x)$ also converges. $\square$