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? 

For any $x>0$, the function
$$ f_x(k)=\frac{1}{(1+kx)^2}$$
is a decreasing function on $\mathbb{R}^+$. It follows that
$$ \sum_{k\geq 1}\frac{1}{(1+kx)^2}=f_x(1)+f_x(2)+f_x(3)+\ldots \leq \int_{0}^{+\infty}f_x(k)\,dk = \frac{1}{x}.$$
The actual value of the series is
$$ \frac{1}{x^2}\cdot\psi'\!\left(1+\frac{1}{x}\right) $$
where $\psi(x)=\frac{d}{dx}\log\Gamma(x)=\frac{\Gamma'(x)}{\Gamma(x)}$ (the $\Gamma$ function is holomorphic in the right half-plane).
