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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?

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$\frac{1}{n^2}$ converges. – quartz Mar 18 '12 at 14:10
$\displaystyle{\sum \frac{1}{n^p}}$ converges if and only if $p > 1$. Note that this is the case in your infinite series. – Sangchul Lee Mar 18 '12 at 14:22
up vote 5 down vote accepted

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.

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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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