<s>[INST] $6n^3+\cos^3n\dots$ 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.
 [/INST] 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).
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