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Question: Prove that the sequence $(x_n)_{n=1}^\infty$ defined by $x_n=\sqrt[3]{n}+1$ tends to infinity.

Definition: $\forall K \in \mathbb{R} \exists N\in \mathbb{N} \forall n \in \mathbb{N} ,n>N:x_n>K $

Proof: Given any $K$ in real numbers, choose $N=(K-1)^3$. Given any $n$ in natural numbers with $n>N=\lceil(K-1)^3\rceil$ we have $x_n=\sqrt[3]{n}+1\geq \sqrt[3]{N}+1\ge K$

Is my proof correct?

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    $\begingroup$ You're looking for \mathbb{N} and \mathbb{R} $\endgroup$ – Alex Kruckman May 11 at 19:28
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    $\begingroup$ Just a small nitpick: your chosen $N$ might not be a natural number. $\endgroup$ – kccu May 11 at 19:31
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    $\begingroup$ The number $N$ can even be negative in this way. I would slightly rephrase the definition: for all $K>0$ there exists $N\in\mathbb{N}$ such that $n\geq N$ implies $x_n\geq K$. $\endgroup$ – RMWGNE96 May 11 at 19:33
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    $\begingroup$ @ViB taking the ceiling function will work! $\endgroup$ – RMWGNE96 May 11 at 19:34
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    $\begingroup$ @ViB in this case, $x_n>0$ for all $n$ so if you can prove the statement for all $K>0$ then you also have proven it for $K\leq 0$. And yes, taking the ceiling function will complete the proof. $\endgroup$ – RMWGNE96 May 11 at 19:45

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