$6n^3+\cos^3n\dots$ Using Comparison Test I have a practice question which I can't figure why the solution comparison is true in the explanation:
$$\frac{6 n^{3}+\cos ^{3} n+5}{5 n^{6}+3 \cos ^{2}(n)+4}$$
Their explaination is:
$-1\leq \cos^3n \leq 1 $ and $ 0 \leq \cos^2n \leq 3$
$\frac{6 n^{3}+\cos ^{3} n+5}{5 n^{6}+3 \cos ^{2}(n)+4} \leq \frac{6 n^{3}+6}{5 n^{6}+4} \leq \frac{12}{5 n^{3}}  $
But I don't understand why I can't just make the comparison:
$\frac{6 n^{3}+\cos ^{3} n+5}{5 n^{6}+3 \cos ^{2}(n)+4} \leq \frac{6}{5 n^{3}}  $
Thanks.
 A: User user has given an answer explaining how the comparison you are making can be made to answer the underlying convergence question, as a limit ratio comparison. I will restrict this answer to an explanation why you can't just make the direct upper-bound comparison that you made.
The practice question's upper bound, $12\over5n^3$, holds for all $n$, whereas your upper bound does not. In fact,
$${6n^3+\cos^3n+5\over5n^6+3\cos^2n+4}\le{6\over5n^3}\implies5n^3\cos^3n+25n^3\le18\cos^2n+24\implies20n^3\le42\implies n\le1$$
so the comparison (for $n\gt1$) actually goes the other way, which makes it useless for proving convergence.
The broader message here is that, in proving things mathematically, the individual, step-by=step assertions you make must be true. There is a useful role for imprecise, even sloppy, thinking when solving mathematical problems (e.g., "Oh, the $6n^3$ and $5n^6$ dominate everything else, so I can throw all the small stuff away"), but at the same time it's important to be able to turn those intuitions into something rigorously true.
A: Your comparison is not an inequality but an asymptotic approximation for $n$ large that is
$$\frac{6 n^{3}+\cos ^{3} n+5}{5 n^{6}+3 \cos ^{2}(n)+4} \sim \frac{6}{5 n^{3}}$$
as $n \to \infty$ indeed
$$\frac{6 n^{3}+\cos ^{3} n+5}{5 n^{6}+3 \cos ^{2}(n)+4}=\frac{6}{5 n^{3}}\frac{1+\frac{\cos ^{3} n}{6n^3}+\frac{5}{6n^3}}{1+3 \frac{\cos ^{2}(n)}{5n^6}+\frac{4}{5n^6}}$$
with
$$\frac{1+\frac{\cos ^{3} n}{6n^3}+\frac{5}{6n^3}}{1+3 \frac{\cos ^{2}(n)}{5n^6}+\frac{4}{5n^6}}\to 1$$

Edit
For the given series
$$\sum_{n=1}^{\infty} \frac{6 n^{3}+\cos ^{3} n+5}{5 n^{6}+3 \cos ^{2} n+4}$$
we can conclude in both ways by the inequality (direct comparison test) or by the asymptotic approximation (limit comparison test).
