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? 

$$x>0\implies(1+kx)^2\ge k^2x^2\implies\frac1{(1+kx)^2}\le\frac1{k^2x^2}$$
and now use the comparison test.
