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? 
 A: $$x>0\implies(1+kx)^2\ge k^2x^2\implies\frac1{(1+kx)^2}\le\frac1{k^2x^2}$$
and now use the comparison test.
A: Hint:
$$\sum_ {k=1}^{\infty} \dfrac{1}{(1+kx)^2}<\sum_ {k=1}^{\infty} \dfrac{1}{(1+k^2x^2)}$$
we know that
$$\sum_ {k=0}^{\infty} \dfrac{1}{(1+k^2x^2)}=\frac{1}{2x}(x+\pi\coth\frac{\pi}{x})$$
A: 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).
A: I wonder Why won't the ratio test be sufficient 
$\displaystyle lim_{n\to\infty} \frac{(n+1)th term}{nth term} = lim_{n\to\infty} \frac{1 + 2kn + n^2*k^2}{1 + 2(k+1)n + (k+1)^2*n^2} = \frac{k^2}{(k+1)^2} <1 $. Hence the series absolutely converges. 
Perhaps this could help you better: https://en.wikipedia.org/wiki/Convergence_tests
