uniform and absolute convergence of $\sum _{n=1}^{\infty }\:\frac{x^n}{1+x^n}$ I'm studying the convergence and absolute convergence of the series of functions defined by the sequence of functions:
\begin{equation*}
f_n: \mathbb{R} \to \mathbb{R},
\end{equation*}
\begin{equation*}
\phantom{1000}x \mapsto \dfrac{x^n}{1 + x^n}
\end{equation*}
I'm trying to bound $\frac{x^n}{1 + x^n}$ by a sequence that its serie is convergent so i could apply Weierstrass M-test but I'm having troubles to find that sequence.
 A: It is clear that
$$\lim_{n\to\infty}\frac{x^n}{1+x^n}=\begin{Bmatrix}0&|x|<1\\\frac{1}{2}&x=1\\1&|x|>1\\DNE&x=-1\end{Bmatrix}$$
thus the series diverges for $|x|\geq 1$. 
The series clearly converges absolutely for $|x|<1$ by the ratio test:
$$\lim_{n\to\infty}\Big|\frac{f_{n+1}(x)}{f_{n}(x)}\Big|=|x|\lim_{n\to\infty}\frac{|1+x^n|}{|1+x^{n+1}|}=|x|<1$$
It remains to see if the series converges uniformly for $|x|<1$. Let's find a bound for the series. We see that $\frac{1}{1+x^n}\leq 1~~,x\geq0$ and we conclude that
$$\sum_{n=0}^{\infty}\frac{x^n}{1+x^n}\leq\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$
Actually, we find that we can do better. Note that $1+x^n\geq \min(1+x,1)$. Then we readily see that
$$|f_n(x)|\leq\frac{|x|^n}{\min(1+x,1)}=M_n(x)$$
and 
$$\sum_{n=0}^{\infty}M_n=\frac{1}{(1-|x|)\min(1+x,1)}<\infty ~~\forall~|x|<1$$
If we restrict the interval to $x\in I_-=(-(1-\delta),0), \delta>0$ or to $x\in I_+=(0,1-\delta), \delta>0$  these bounds allow us to prove uniform convergence. Notice that $|x|<1-\delta$ and $1+x>\delta$ which allow us to show that in $I_+$;
$$|f_n(x)|\leq(1-\delta)^n=\bar{M}_n$$
$$\sum_{n=0}^{\infty}\bar{M}_n=\frac{1}{\delta}$$
and in $I_-$ respectively
$$|f_n(x)|\leq\frac{(1-\delta)^n}{\delta}=\bar{M}_n$$
$$\sum_{n=0}^{\infty}\bar{M}_n=\frac{1}{\delta^2}$$
so the intervals separately satisfy the M-test for uniformity.
