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Let $(x_n)_{n \geq 1}$ be a sequence with $x_1=1$, defined by \begin{align*} x_{n+1}=x_n+\frac{1}{3x_n^2} \end{align*} Prove that there exists a fixed $c>0$ such that $x_n<\sqrt[3]{n+c}$

I cubed the relation in order to prove that $x_n \geq \sqrt[3]{n}$, and I guess that $c$ may be $1$, but I couldn't prove it.


marked as duplicate by Martin R, Community Jan 7 '17 at 20:43

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  • $\begingroup$ First I tried induction, but failed. Then I tried using a similar approach as for the "$\geq$" part, but I ended up with a sum in the form of 1+1/2+...+1/n, which doesn't converge to any constant. $\endgroup$ – Shroud Jan 7 '17 at 20:13
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    $\begingroup$ As shown in the "possible duplicate", the statement is wrong. $\endgroup$ – Martin R Jan 7 '17 at 20:34