# Prove that the sequence is convergent

How can we show that the sequence $$a_n=\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3-n^2}$$ is convergent?

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HINT

The sequence converges.

Use the identity $$a- b = \dfrac{a^3 - b^3}{a^2 + b^2 + ab}$$

$a_n =\sqrt[3]{n^3+n^2} - \sqrt[3]{n^3-n^2}$, so when $n$ is large enough, \begin{align} a_n &= n\cdot \sqrt[3]{1+\frac{1}{n}} - n\cdot \sqrt[3]{1 - \frac{1}{n}} \\ &= n\cdot \left(1+\frac{1}{3n} - \frac{1}{9n^2} + \mathcal{O}(n^{-3})\right) - n\cdot \left(1 - \frac{1}{3n} - \frac{1}{9n^2} + \mathcal{O}(n^{-3})\right) \\ &= \frac23 + \mathcal{O}(n^{-2}) \end{align} It seems that $a_n$ converges to $\frac23$.