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Channel: Convergence of $\sum_{n=0}^{\infty} \left(\sqrt[3]{n^3+1} - n\right)$ - Mathematics Stack Exchange
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Answer by John Molokach for Convergence of $\sum_{n=0}^{\infty}...

\begin{align*}\sqrt[3]{n^3+1}-n&=\frac{\left(\sqrt[3]{n^3+1}-n\right)\left(\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)}+n^2 \right )}{\left(\sqrt[3]{(n^3+1)^2}+n\sqrt[3]{(n^3+1)}+n^2 \right...

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Answer by Jack D'Aurizio for Convergence of $\sum_{n=0}^{\infty}...

For any $x\in[0,1]$, the inequality:$$ (1+x)^{\frac{1}{3}}\leq 1+\frac{x}{3} \tag{1}$$holds by concavity or just by considering the cube of both terms. It follows that:$$ 0\leq \sqrt[3]{1+n^3}-n =...

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Answer by Travis Willse for Convergence of $\sum_{n=0}^{\infty}...

Hint We can apply a cubic analogue of the technique usually called multiplying by the conjugate, which itself is probably familiar from working with limits involving square roots. In this case, write...

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Answer by Nikita Evseev for Convergence of $\sum_{n=0}^{\infty}...

Hint.Note $\sqrt[3]{n^3+1} = n\sqrt[3]{1+1/n^3}$,and then $\sqrt[3]{1+1/n^3} = 1 + \frac{1}{3n^3} + o(\frac{1}{n^3})$.Next, I belive, limit comparison test will work.

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Convergence of $\sum_{n=0}^{\infty} \left(\sqrt[3]{n^3+1} - n\right)$

I have encountered the following problem:Determine whether $$\sum \limits_{n=0}^{\infty} \left(\sqrt[3]{n^3+1} - n\right)$$ converges or diverges.What I have tried so far:Assume that $a_n =...

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