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...
View ArticleAnswer 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:$$ 0\leq \sqrt[3]{1+n^3}-n =...
View ArticleAnswer by Travis Willse for Convergence of $\sum_{n=0}^{\infty}...
Hint We can apply a cubic analogue of the technique usually called multiplying by the conjugate, which itself is probably familiar from working with limits involving square roots. In this case, write...
View ArticleAnswer by Nikita Evseev for Convergence of $\sum_{n=0}^{\infty}...
Hint.Note $\sqrt[3]{n^3+1} = n\sqrt[3]{1+1/n^3}$,and then $\sqrt[3]{1+1/n^3} = 1 + \frac{1}{3n^3} + o(\frac{1}{n^3})$.Next, I belive, limit comparison test will work.
View ArticleConvergence of $\sum_{n=0}^{\infty} \left(\sqrt[3]{n^3+1} - n\right)$
I have encountered the following problem:Determine whether $$\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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