Proving uniform convergence on an unbounded interval 
Prove that the following series 
  $$\sum_{n=1}^{\infty}\frac{nx}{1+n^2\log^2(n)x^2}$$
  converges uniformly on $[\epsilon,\infty)$ for any $\epsilon>0.$

What I have done:
The function 
$$f_n(x)=\frac{nx}{1+n^2\log^2(n)x^2}$$ is a decreasing function on $[1,\infty)$. So on $[1,\infty)$ its  maximun value is 
$$f_n(1)=\frac{n}{1+n^2\log^2(n)}\leq\frac{n}{n+n^2\log^2(n)}=\frac{1}{1+n\log^2(n)}=M_n.$$
By Cauchy Condensation Test, $\sum M_n$ is convergent. Therefore by the Weierstrass-M Test $\sum f_n(x)$ is convergent on $[1,\infty)$. 
I am not entirely sure about the following statement. Please correct me if I am wrong: 


If we can prove that the $\sum f_n(x)$ is  uniformly convergent on both $[\epsilon,1]$ and $[1,\infty)$, then it is uniformly convergent on $[\epsilon, \infty)$.   


But I don't know how to prove that $\sum f_n(x)$ is uniformly convergent on $[\epsilon,1]$. I need some help.

The function $f_n$ attains its maximum when $x=\frac1{n\log n}$, if $\frac {1}{n\log n}<\varepsilon$; its value there is $\frac1{\log n}$. Of course, if $n$ is large enough, $\frac{1}{n\log n}\leqslant\varepsilon$ and so the maximum of $f_n$ will be then $f_n(\varepsilon)$. But$$f_n(\varepsilon)=\frac{n\varepsilon}{1+n^2\log^2(n)\varepsilon^2}<\frac1{n\log^2(n)\varepsilon}.$$As you wrote, we can use the Cauchy condensation test here.
