The series $\sum\limits_ {k=1}^{\infty} \frac{1}{(1+kx)^2}$ converges for $x>0$ $$ \sum_ {k=1}^{\infty} \dfrac{1}{(1+kx)^2}$$  $$x\in (0,\infty) $$ 
This series converges on given interval but how exactly can I show this is true? 

Hint:
$$\sum_ {k=1}^{\infty} \dfrac{1}{(1+kx)^2}<\sum_ {k=1}^{\infty} \dfrac{1}{(1+k^2x^2)}$$
we know that
$$\sum_ {k=0}^{\infty} \dfrac{1}{(1+k^2x^2)}=\frac{1}{2x}(x+\pi\coth\frac{\pi}{x})$$
