Does $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^2}{(1+x^2)^n}$ uniformly converge?

I want to use Leibniz test to check uniformly convergence $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^2}{(1+x^2)^n}$. In order to do that I need to check whether the series $\frac{(-1)^{n-1}x^2}{(1+x^2)^n}$ uniformly converges to 0. It's easy to check that it is pintwiseconverges to 0, I'd like your help in deciding if the convergence is uniform. I thought of checking the $\lim_{n \to \infty} |f_n(x)-f(x)|$, I tried to derive $r_n(x)=f_n(x)-f(x)$, I got that $x_1=o, x_2=\frac{1}{n-1}$, both donate that the lim $r_n(x)$ is 0 when $n \to \infty$, Can I use that to conclude that the function does uniformly converges?

the radius of $x$ is all the real line.

Thanks a lot!

-
Use that given an $\epsilon>0$ you can choose $n$ large enough that $\frac{x^2}{(1+x^2)^n}$ will be smaller than $\epsilon$ regardless of the value of x. Then use that the Leibniz test gives you a bound on any given partial sum minus the limit function ( en.wikipedia.org/wiki/Alternating_series_test - I assumed this is mentioned in your textbook). – testcase Feb 13 '12 at 19:29

The $N$th rest of the series is $R_N(x)=\displaystyle\sum_{n=N+1}^{\infty}\frac{(-1)^{n-1}x^2}{(1+x^2)^n},$ hence $$R_N(x)=\frac{(-1)^Nx^2}{(1+x^2)^{N+1}}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(1+x^2)^n}=\frac{(-1)^Nx^2}{(1+x^2)^{N}(2+x^2)},$$ and $\left|R_N(x)\right|\leqslant U_N\left(\frac{x^2}{1+x^2}\right)$ with $U_N(t)=t(1-t)^N$ for every $t$ in $[0,1]$. The function $U_N$ achieves its maximum at $t_N=\frac1{N+1}$ hence $$\|R_N\|_{\infty}\leqslant U_N(t_N)=\frac{N^N}{(N+1)^{N+1}}.$$ Using the upper bound $\left(\frac{N}{N+1}\right)^{N+1}\leqslant\frac1{\mathrm e}$, one gets finally $\|R_N\|_{\infty}\leqslant\frac1{\mathrm eN}\to0$.