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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.

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

2 Answers 2

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By limit comparison test we have that

$$\frac{\frac{n^5+1}{6n^{10}+\sqrt[6]{n}}}{\frac1{n^4}}=\frac{n^9+n^4}{6n^{10}+\sqrt[6]{n}} \to 0$$

therefore the given series converges.

As an alternative by direct comparison test we have

$$\frac{n^5+1}{6n^{10}+\sqrt[6]{n}} \le \frac{n^5+n^5}{6n^{10}+0}=\frac13\frac1{n^5}$$

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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.

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