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?