Comparison test for $\sum_{n=1}^{\infty} \frac{\sqrt{n^5 + n^3 - n}}{2n^3 + 3n^2 + 1}$

$$\frac{\sqrt{n^5 + n^3 - n}}{2n^3 + 3n^2 + 1}\ge\frac{\sqrt{n^5}}{2n^3+2n^3+2n^3}=\frac{n^{5/2}}{6n^3}=\frac16\frac1{\sqrt n}$$

and since $\;\sum\frac1{\sqrt n}\;$ diverges ...

Yes, it is permissible, and instead of worrying about a direct comparison you should use the limit comparison test: since $$\lim_{n \to \infty} \frac{\dfrac{\sqrt{n^5+n^3-n}}{2n^3 + 3n^2 + 1}}{\dfrac{\sqrt{n^5}}{2n^3}} = 1$$ the series $$\sum_{n=1}^\infty \dfrac{\sqrt{n^5+n^3-n}}{2n^3 + 3n^2 + 1} \quad \text{and} \quad \sum_{n=1}^\infty \dfrac{\sqrt{n^5}}{2n^3}$$ either both converge or both diverge.

It is much simpler to use this result from asymptotic analysis:

Let $\sum_n u_n$ and $\sum_n v_n$ two series with (ultimately) positive terms, such that $u_n\sim_\infty v_n$. Then $\sum _n u_n$ and $\sum_n v_n$ both converge or both diverge.

Now a polynomial is asymptotically equivalent to its leading term, so $$\frac{\sqrt{n^5 + n^3 - n}}{2n^3 + 3n^2 + 1}\sim_\infty\frac{\sqrt{n^5}}{2n^3}=\frac 1{2\sqrt n},$$ and the latter is a divergent $p$-series.