Determine whether the series is convergent or divergent: $\sum_{n=1}^\infty \frac{n+5^{}}{{(n^{7}+n^{2})}^{1/3}}$

Question

I am confused as to what test I should use to determine whether the series below is convergent or divergent.

$$\sum_{n=1}^\infty \frac{n+5^{}}{{(n^{7}+n^{2})}^{1/3}}$$

Answer 1

Consider $a_{n} = \frac{n+5}{(n^{7}+n^{2})^{1/3}}$ and $b_{n} = \frac{1}{n^{4/3}}$

Then,

\begin{align*} \lim_{n\rightarrow \infty} |a_{n}/b_{n}| = \lim_{n\rightarrow\infty} \frac{(n+5)n^{4/3}}{(n^{7}+n^{2})^{1/3}} = \lim_{n\rightarrow\infty} \frac{n^{7/3}}{n^{7/3}} = 1 \end{align*}

Since $\lim_{n\rightarrow \infty} |a_{n}/b_{n}|$ is finite and non zero, $\sum a_{n}$ and $\sum b_{n}$ must converge or diverge together, but $b_{n}$ is a convergent $p$ series.

Answer 2

Before trying any tests for convergence/divergence, it's important to have a gut feeling about the series. Here's how you could do it here: Upstairs you have $n+5.$ The $5$ is so small compared with large $n$ that you should ignore it. Similarly, the $n^2$ downstairs is way smaller than $n^{7}.$ Ignore that too. You're left with $n/n^{7/3} = 1/n^{4/3}.$ Since $\sum 1/n^{4/3} <\infty,$ your thought has to be the original series converges. So there's you gut feeling. Now go prove it using the tests!