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Does this series converge or diverge?

$$\sum_{n=1}^\infty\dfrac{n^2-3n}{\sqrt[3]{n^{10}-4n^2}}$$

I have tried using the comparison test, however when simplifying this i get $1/\sqrt[3]n$. I do not know what to compare with, I have tried comparing with something related with the p-test but it does not work.

Any ideas?

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    $\begingroup$ Yes, the series converge. $\endgroup$ – Hakim Apr 16 '14 at 13:23
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    $\begingroup$ Yes, thanks, badly (wrongly) put. $\endgroup$ – André Nicolas Apr 16 '14 at 13:39
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$\frac{n^2-3n}{(n^{10}-4n^2)^{\frac{1}{3}}}\sim_{+\infty}\frac{n^2}{n^{10\frac{1}{3}}}=\frac{n^2}{n^{15}}=\frac{1}{n^{\frac{4}{3}}}$ hence the series converges.

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$$\frac{n^2-3n}{\sqrt[3]{n^{10}-4n^2}}\sim\frac{n^2}{\sqrt[3]{n^{10}}}=\frac{n^2}{n^{10/3}}=\frac1{n^{4/3}}$$

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We have that $$ n^2-3n\sim n^2 $$ and $$ \sqrt[3]{n^{10}-4n^2}\sim n^{10/3}, $$ where $\sim$ denotes that the ratio of the two sides goes to $1$ as $n\to\infty$. Hence, $$ \frac{n^2-3n}{\sqrt[3]{n^{10}-4n^2}}\sim n^{-4/3} $$ and the series converges.

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Notice that for $n$ large enough, we have the following inequality $$0<\frac{n^2-3n}{\sqrt[3]{n^{10}-4n^2}}<\frac{n^2}{n^{\frac{10}{3}}\sqrt[3]{1-4/n^{8}}}<\frac{n^2}{n^{\frac{10}{3}}\sqrt[3]{1/2}}=\frac{\sqrt[3]{2}}{n^{\frac{4}{3}}}.$$ That could help.

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