<s>[INST] 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? 
 [/INST] your general term is equivalent to $1/n^2$, so it converges.
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