Technique for using direct and limit comparison tests?

Question

Suppose you have an infinite series $$\sum_{n=1}^{\infty} \frac{1}{n^2 + 9}$$

Using the direct comparison test, the sequence can obviously be compared to $\sum_{n=1}^\infty\frac{1}{n^2}$ since it is smaller, and then by the p-series test, since $\sum\frac{1}{n^2}$ converges $\sum\frac{1}{n^2+9}$ also converges. This makes sense since when $\lim{n\to \infty}$, the $9$ is negligible.

In other cases though, the series is not simple. Take for example: $$\sum_{n=1}^\infty\frac{\ln^4(x)}{n}$$ The end behavior at $\infty$ is not as simple as the first one. How do you know what sequence to compare $\frac{\ln^4(n)}{n}$ to? How do you know what term is dominant? Take the derivative and compare? Is there a technique?

Another example: $$\sum_{n=2}^\infty\frac{n}{n^3-1}$$ I know the sequence compares to $\frac{1}{n^2}$, but is there some logical technique that allows us to make that comparison?

Essentially, my question is how do you figure out the comparison sequence? Before, I have been taught to take the most dominant term in the numerator and denominator and use those as a comparison, but I feel this technique is prone to error. I realize these questions are a bit elementary, but I think the basic logic would be useful.

Thank you.

Answer 1

Alternative method: $$\sum_{n=1}^{\infty} \frac{n}{n^3 - 1}=\sum_{n=1}^{\infty} \frac{1/n^2}{1 - 1/n^3} \rightarrow 0$$

Answer 2

Excuse me, but $log^4n \geq 1$ for $n \geq 3$, so $\sum \frac{log^4 n}{n}$ is bigger than $\sum \frac{1}{n}$, which diverges. Did you get it? Basically, you learn about simple sequences, then you learn about simples series and after that, you deal with series more complicated. The same you do with integrals, you know what I mean?