This is problem 9.3.1 from Bruckner's 'Elementary real Analysis':
I've found previously that it is uniform convergent in $(0,1)$, but it seems to me that for an appropriate $a>0$, this sequence is also uniformly convergent in $(- \infty,-a)$ and $(a, \infty)$, to the function $f(x) = 1$ since $\lim_{x\to\pm\infty} f_n(x)=1$. However I don't know how con one determine $a$, and then show it is uniformly convergent on those intervals.
On
Hint
$$f_n(x)=1-\frac{1}{1+x^n}$$
so, $$x>1\implies \lim_{n\to+\infty}f_n(x)=1$$ because $\lim x^n=+\infty$.
$$x=1\implies \lim_{n\to+\infty}f_n(x)=\frac 12$$
$$-1<x<1\implies \lim_{n\to+\infty}f_n(x)=0$$ because $\lim x^n=0$.
finally $$x<-1\implies \lim_{n\to+\infty}f_n(x)=1$$
there is a pointwise convergence at $ (-\infty,-1)\cup(-1,+\infty)$. You can check easily that there is a uniform convergence at $ [a,+\infty)$ for any $ a>1$ since
$$\sup_{x\ge a}|f_n(x)-1|=\frac{1}{1+a^n}$$
If a sequence of functions converges uniformly it converges pointwisely and the limit is the same, so the first thing you should do is to compute the eventual pointwise limit. Here you may notice that for odd $n$, $f_n(-1)$ is not defined so the comon domain of your functions is $D = (-\infty,-1)\cup (-1,\infty)$. There is pointwise convergence on $D$ to $$ f(x) = \begin{cases} 1, \ x> 1 \\ 1/2, \ x = 1 \\ 0, \ -1 < x < +1 \\ 1, \ x < -1 \end{cases} $$ that is not continuous. As a uniform limit of continuous functions is continuous, you cannot have uniform convergence on the whole $D$. Moreover on an open interval containing $1$ you cannot have uniform convergence because you have a discontinuity point at $1$ for $f$. Same thing at $-1$, so the only reasonable sets to check if the sequence converges uniformly are $(-\infty,-1)$, $(-1,1)$, $(1,\infty)$ or any compact interval of the last sets.
Now choose some interval $I$ as explained, fix $n$ and compute by hand $$ \sup_{x \in I} |f(x) - f_n(x)|. $$ To do so you can inspect the variations of $f-f_n$.