Spivak ed.1 Chapter 22 problem 1 (v)

Question

The problem asks to determine whether the following series is convergent of not: $$ \sum_{n=2}^\infty \frac{1}{\sqrt[3]{n^2-1}} $$ The solution states that it is convergent because for very big n values, $$ \frac{1}{\sqrt[3]{n^2-1}} \lt \frac{2}{\sqrt[3]{n^2}} $$ But I don't get why $\frac{2}{\sqrt[3]{n^2}}$ is convergent, I have solved this way: $$ n^2-1 \lt n^3 \ \ \forall n\gt 2$$ $$ \sqrt[3]{n^2-1} \lt n$$ $$ \frac{1}{n} \lt \frac{1}{\sqrt[3]{n^2-1}}$$ And since $ \sum_{n=2}^\infty \frac{1}{n}$ is divergent, $\sum_{n=2}^\infty \frac{1}{\sqrt[3]{n^2-1}}$ should be too. Could you help me out?