Comparison test for $\sum_{n=1}^{\infty} \frac{\sqrt{n^5 + n^3 - n}}{2n^3 + 3n^2 + 1}$ I'm trying to use comparison test to show that this series diverges:
$$\sum_{n=1}^{\infty} \frac{\sqrt{n^5 + n^3 - n}}{2n^3 + 3n^2 + 1}$$
Here are the steps I have come up with but I am not sure if it is right.
$$\sum_{n=1}^{\infty} \frac{\sqrt{n^5}}{3n^3}\ \text{diverges (by integral test)}$$
$$0 < \frac{\sqrt{n^5}}{3n^3} \leq a_n\ \text{(for large n)}$$
$$\therefore a_n\ \text{diverges}$$
Note: I added "for large n" since according to Wolfram|Alpha, it is only true for n > 2.671... is this permissible or does it mean I have made an error?
 A: $$\frac{\sqrt{n^5 + n^3 - n}}{2n^3 + 3n^2 + 1}\ge\frac{\sqrt{n^5}}{2n^3+2n^3+2n^3}=\frac{n^{5/2}}{6n^3}=\frac16\frac1{\sqrt n}$$
and since $\;\sum\frac1{\sqrt n}\;$ diverges ...
A: 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.
A: 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.
