Uniform convergence $\sum_{n=1}^{\infty} \frac{x^2}{(1+x^2)^n}$ if $|x| > c$ for all $c>0$

Question

Problem : Prove $\sum_{n=1}^{\infty} \frac{x^2}{(1+x^2)^n}$ : converges uniformly if $|x| > c$ for all $c>0$

I'm stuck.. Can anybody provide me a hint to solve this..?

My attempt : Choose d s.t. $|x| > d$ and

Answer 1

$$f_n(x)=\sum_{n=1}^{\infty}\dfrac{x^2}{(1+x^2)^n}=\sum_{n=1}^{\infty}x^2\dfrac{1}{(1+x^2)^n}$$ is a geometric series. And it will converge if $\dfrac{1}{(1+x^2)^n}<1\implies x\neq0$.

For all $x\neq0$, using the formula for geometric series, $\sum_{n=1}^{\infty}\dfrac{x^2}{(1+x^2)^n}=1+x^2,x\neq0$, therefore $f_n(x)$ converges poinwise to $f$ when $x\neq0$. Use the definition of uniform convergence in terms of suprema: $$\lim_{n\to\infty}|f(x)-f_m(x)|=\lim_{n\to\infty}\left| 1+x^2-x^2 \frac{1-(1+x^2)^{-(m+1)}}{1-(1+x^2)^{-1}}\right|=0=\lim\sup_{x\neq0}|f(x)-f_m(x)|$$

Answer 2

Weiestraß' $M$-test: Use that for $x$ with $|x|>c$ we have $$\frac{x^2}{(1+x^2)^n}\leq \frac{x^2}{1+x^2}\frac{1}{(1+c^2)^{n-1}}\leq \frac{1}{(1+c^2)^{n-1}}.$$