TITLE: If the limit comparison test fails
QUESTION [0 upvotes]: Just want to make sure I'm applying the limit comparison test correctly. And if so and it does indeed fail, what should be my next step in determining if the series diverges?
$$\sum_{n=1}^\infty \frac{5n^3 + 7n^2 + 9}{n^6+2n^3+1}$$
Taking the dominant terms this reduces to $\frac{1}{n^3}$ which converges.
$$\frac{5n^3 + 7n^2 + 9}{n^6+2n^3+1} * \frac{n^3}{1}$$
$$= \frac{5n^6 + 7n^5 + 9n^3}{n^6+2n^3+1}$$
$$= \frac{5 + 7n^5 + 9}{2+1}$$
$$= \frac{7n^5 + 14}{3}$$
$\frac{7n^5 + 14}{3}$ isn't finite for $n=\infty$, so the test fails. Where should I go from here?

REPLY [1 votes]: After this
$$ \frac{5n^6 + 7n^5 + 9n^3}{n^6+2n^3+1}$$
you made errors.
The limit above is finite and it's values is $5$
Indeed
$$\lim_{n\to\infty}\frac{5n^6 + 7n^5 + 9n^3}{n^6+2n^3+1}=\lim_{n\to\infty}\frac{n^6\left(5+\frac{7}{n}+\frac{9}{n^3}\right)}{n^6\left(1+\frac{2}{n^3}+\frac{1}{n^6}\right)}=\lim_{n\to\infty}\frac{5+\frac{7}{n}+\frac{9}{n^3}}{1+\frac{2}{n^3}+\frac{1}{n^6}}=5$$
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