Convergence of $\sum_{n=0}^{+ \infty} \frac{n^3 - 5n^2 + \pi}{2n^5 - \sqrt{3}}$ 
Does $$\sum_{n=0}^{+\infty} \frac{n^3 - 5n^2 + \pi}{2n^5 - \sqrt{3}}$$
converge?

My attempt:
$$\forall n \in \mathbb{N} \setminus\{0\}:\frac{n^3-5n^2 + \pi }{2n^5 -\sqrt{3}} = \frac{1}{n^2}\frac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5}$$
And because 
$$\frac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5} \to \frac{1}{2}$$ 
there exists $n_0$ such that $\forall n > n_0: \left\vert \dfrac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5}  - \dfrac{1}{2}\right\vert < 1$
Hence, for $n \geq n_0$:
$$ \frac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5} < \frac{3}{2}$$
Thus: $$\frac{n^3-5n^2 + \pi }{2n^5 -\sqrt{3}} < \frac{3}{2}\frac{1}{n^2}$$
and for $n_1$ sufficiently large, the sequence of terms is positive (as both the numerator and the denumerator converge to $+ \infty$, so for $n$ large enough the numerator and denumerator are positive, and hence the quotient is positive). 
Let $N :=  \max\{n_0,n_1\}$
Then, because $\sum\dfrac{1}{n^2}$ converges, we conclude using the comparison test that the series 
$$\sum_{n > N}^{+ \infty} \frac{n^3-5n^2 + \pi }{2n^5 -\sqrt{3}}$$
converges, and hence, the given series converges.
Is this correct? (I tried to explain every single step). Do you have any comments? Is there an easier approach? 

your general term is equivalent to $1/n^2$, so it converges.
