Sequence tending to infinity (checking epsilon proof)

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?

• You're looking for \mathbb{N} and \mathbb{R} – Alex Kruckman May 11 at 19:28
• Just a small nitpick: your chosen $N$ might not be a natural number. – kccu May 11 at 19:31
• 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 $x_n\geq K$. – RMWGNE96 May 11 at 19:33
• @ViB taking the ceiling function will work! – RMWGNE96 May 11 at 19:34
• @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. – RMWGNE96 May 11 at 19:45