# Comparison test to check the nature of a series

Reading my calculus notes I found the next example:

To analyze the nature of the series $$S=\sum_{n=1}^{\infty} \frac{n^5+1}{6n^{10}+\sqrt[6]{n}}$$ by the limit comparison test, a series that can be compared to $$S$$ is $$\sum_{n=1}^{\infty}\frac{1}{n^4}$$

I don't understand how to arrive to such conclusion. Any help is appreciated.

• Which conclusion? That $\sum\frac1{n^4}$ is a good candidate or that applying the test actually works? Commented Oct 9, 2020 at 8:44

$$\frac{\frac{n^5+1}{6n^{10}+\sqrt[6]{n}}}{\frac1{n^4}}=\frac{n^9+n^4}{6n^{10}+\sqrt[6]{n}} \to 0$$
$$\frac{n^5+1}{6n^{10}+\sqrt[6]{n}} \le \frac{n^5+n^5}{6n^{10}+0}=\frac13\frac1{n^5}$$
The shortest method would use asymptotic equivalence of functions. Namely as any polynomial is equivalent to its leading term, we have $$n^5+1\sim_\infty n^5,\quad 6n^{10}+\sqrt[6]{n}\sim_\infty 6n^{10}, \quad\text{ therefore }\quad \frac{n^5+1}{6n^{10}+\sqrt[6]{n}}\sim_\infty \frac{n^5}{6n^{10}}=\frac1{6n^5},$$ which is a convergent $$p$$-series.