<s>[INST] Convergence of $\sum_{n=1}^{\infty} \frac{n+5}{\sqrt[3]{n^7+n^2-n}}$ Convergence of 
$$\sum_{n=1}^{\infty} \frac{n+5}{\sqrt[3]{n^7+n^2-n}}$$
I think ratio test will be very tedious, root test too since its not a exponential equation. So I tried limit comparison test, noting that 
$$\frac{n+5}{\sqrt[3]{n^7+n^2-n}} \approx \frac{n}{n^{7/3}} \approx \frac{1}{n^{4/3}} \approx \frac{1}{n^2}, \qquad \frac{1}{n^2} \text{ diverges}$$
$$\lim_{n\to\infty} \frac{n+5}{\sqrt[3]{n^7 + n^2 - n}} \cdot n^2 = \lim_{n\to\infty} \frac{n^3+5n^2}{\sqrt[3]{n^7 + n^2 - n}}$$
Now, should I multiply top & bottom by $\frac{1}{n^3}$? But will the bottom become 
$$= \lim_{n\to\infty} \frac{1+\frac{5}{n}}{\sqrt[3]{\frac{n^7 + n^2 - n}{n^{27}}}} = \frac{1}{0}$$
Which seems wrong? 
 [/INST] Seems a straightforward comparison test will also work.
$$
0\le \sum_1^\infty \frac{n+5}{\sqrt[3]{n^7 +n^2 - n}}
\le \sum_1^\infty \frac{n+5n}{\sqrt[3]{n^7}} = 6\sum_1^\infty \frac{1}{n^\frac43}.
$$
On the right we have a convergent p-series. The second inequality is valid because $5\le 5n$ and $n^2-n\ge 0$ for $n\ge 1$.
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