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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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2 Answers 2

up vote 3 down vote accepted

HINT

The sequence converges.

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

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\begin{align}\sqrt[3]{n^3 + n^2} - \sqrt[3]{n^3 - n^2} & = \dfrac{(n^3 + n^2)- (n^3-n^2)}{(n^3+n^2)^{2/3} + (n^3-n^2)^{2/3} + (n^6-n^4)^{1/3}}\\ & = \dfrac{2n^2}{n^2 \left(\left(1+1/n \right)^{2/3} + \left(1-1/n \right)^{2/3} + (1-1/n^2)^{1/3}\right)}\\& = \dfrac2{\left(1+1/n \right)^{2/3} + \left(1-1/n \right)^{2/3} + (1-1/n^2)^{1/3}}\end{align}

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@ Marvis I tried already,but could not proceed –  p.s Oct 24 '12 at 23:01
    
@Pilot: It's not useless. –  wj32 Oct 25 '12 at 0:09

$a_n =\sqrt[3]{n^3+n^2} - \sqrt[3]{n^3-n^2}$, 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 $\frac23$.

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