Series convergence with parameter: 
Determine whether the following series :
  $$\sum_{n=2}^\infty \frac {1+xn}{\sqrt{n^2+n^6x}}, x \in \mathbb R^+_0$$
  converges absolutely, conditionally or diverges.

I tried to estimate the series for $n > x$ using the following:
$$\sum_{n=2}^\infty \frac {1+xn}{\sqrt{n^2+n^6x}} \leq \sum_{n=2}^\infty \frac {2xn}{\sqrt{n^6x}} = 2\sqrt x\sum_{n=2}^\infty n^{-2}$$
Which would mean that the series is absolutely convergent for every $x \in \mathbb R^+_0$?
Is this the correct way to go about this or am I overlooking something?
 A: Note that the series diverges if $x = 0$.  
You need a slight correction to your argument. For fixed $x > 0$, we have for all $n > 1/x$,
$$ \frac{1+xn}{\sqrt{n^2 + n^6x}} \leqslant \frac{2nx}{n^3\sqrt{x}} = \frac{2\sqrt{x}}{n^2},$$
and by the comparison test we have pointwise convergence for all $x > 0$.
The convergence is not uniform for $x \in (0,\infty)$.  Note that
$$\sup_{x \in [0,\infty)}\sum_{n=m+1}^{\infty} \frac{1+xn}{\sqrt{n^2 + n^6x}} >\sup_{x \in [0,\infty)}\sum_{n=m+1}^{2m} \frac{1}{n\sqrt{1 + n^4x}}  > \sup_{x \in [0,\infty)}\frac{m}{(2m)\sqrt{1+(2m)^4x}} \\ =  \sup_{x \in [0,\infty)}\frac{1}{2\sqrt{1+16m^4x}}   \geqslant \underbrace{\frac{1}{2\sqrt{1+16m^4{m^{-4}}}}}_{\text{ taking } x = m^{-4}} = \frac{1}{2\sqrt{17}}  $$
and the RHS does not converge to $0$ as $m \to \infty$.
The convergence is uniform on any compact interval $[a,b]$ with $a > 0$. I will leave this for you to prove.
A: Perhaps you meant $n > \frac{1}{x}$? Then 
$$
\sum_{n=2}^{\infty} \frac{1+xn}{\sqrt{n^2 + n^6 x}} \leq 
\sum_{n=2}^{\infty} \frac{2nx}{\sqrt{n^6 x}} =
\sum_{n=2}^{\infty} \frac{2 \sqrt{x}}{n^2} =
2\sqrt{x} \sum_{n=2}^{\infty} \frac{1}{n^2}
$$
which is convergent for each $x>0$.
Edit: After I posted my answer I saw that @RRL had posted a more complete answer.
