Does the series converge or diverge?$\sum_{n=1}^\infty\frac{n^2-3n}{\sqrt{n^{10}-4n^2}}$

Does this series converge or diverge?

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

I have tried using the comparison test, however when simplifying this i get $1/\sqrtn$. 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?

• Yes, the series converge. – Hakim Apr 16 '14 at 13:23
• Yes, thanks, badly (wrongly) put. – André Nicolas Apr 16 '14 at 13:39

$\frac{n^2-3n}{(n^{10}-4n^2)^{\frac{1}{3}}}\sim_{+\infty}\frac{n^2}{n^{10\frac{1}{3}}}=\frac{n^2}{n^{15}}=\frac{1}{n^{\frac{4}{3}}}$ hence the series converges.
$$\frac{n^2-3n}{\sqrt{n^{10}-4n^2}}\sim\frac{n^2}{\sqrt{n^{10}}}=\frac{n^2}{n^{10/3}}=\frac1{n^{4/3}}$$
We have that $$n^2-3n\sim n^2$$ and $$\sqrt{n^{10}-4n^2}\sim n^{10/3},$$ where $\sim$ denotes that the ratio of the two sides goes to $1$ as $n\to\infty$. Hence, $$\frac{n^2-3n}{\sqrt{n^{10}-4n^2}}\sim n^{-4/3}$$ and the series converges.
Notice that for $n$ large enough, we have the following inequality $$0<\frac{n^2-3n}{\sqrt{n^{10}-4n^2}}<\frac{n^2}{n^{\frac{10}{3}}\sqrt{1-4/n^{8}}}<\frac{n^2}{n^{\frac{10}{3}}\sqrt{1/2}}=\frac{\sqrt{2}}{n^{\frac{4}{3}}}.$$ That could help.