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? 
 A: You almost did the right thing at the start. 
Compare with the series whose terms are the "important' terms in the expression $$  {\color{maroon}n+5\over \root 3\of {\color{maroon}{n^7}+n^2-n}}. $$ 
This leads you to compare with the series $$\sum\limits_{n=1}^\infty {1\over n^{4/3}}.$$
(do not compare with the series whose terms are $1/n^2$).
Note that  $\sum\limits_{n=1}^\infty {1\over n^{4/3}}$ is a convergent $p$ series 
(the $p$-series $\sum\limits_{n=1}^\infty  {1\over n^p}$ converges if and only if $p>1$).
 So the original series will converge, if the Limit  Comparison Test applies. But, of course,
we need to check if the Limit Comparison Test applies; so, we take the limit 
$$\eqalign{ \lim_{n\to\infty} 
{{ n+5\over \root 3\of {n^7+n^2-n} }\over {1\over n^{4/3}}}
&=
\lim_{n\to\infty} 
{{ n+5\over \root 3\of {n^7+n^2-n} }\cdot  {  n^{4/3}}}\cr
&=
\lim_{n\to\infty} 
{{ n^{7/3}+5n^{4/3}\over \root 3\of {n^7(1+{1\over n^5}-{1\over n^6})} } }\cr
&=
\lim_{n\to\infty} 
{{ n^{7/3}(1+ {5\over {n^{3/3}}})\over 
  n^{7/3}\root 3\of {1+{1\over n^5}-{1\over n^6}} } }\cr
&=
\lim_{n\to\infty} 
{{ 1+ {5\over {n^{3/3}}} \over 
  \root 3\of {1+{1\over n^5}-{1\over n^6}} } }\cr
&=1.
}
$$
Since limit value is 1, the Limit  Comparison Test indeed applies; and thus $\sum\limits_{n=1}^\infty   { n+5\over \root 3\of {n^7+n^2-n} }$ is a convergent series.
A: 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 $n\ge 1$.
