# Prove $\forall K > 0: \lim_{n\rightarrow\infty} \sqrt[n]{K} = 1$

Alright, so I've already proven that both $\forall n \in \mathbb{N}:\lim_{n\rightarrow\infty}\sqrt[n]{n} = 1$ and $\forall K\geq 1:\lim_{n\rightarrow\infty}\sqrt[n]{K}=1$.

I got the feeling, that I can prove $\forall K \in \mathbb{R}> 0: \lim_{n\rightarrow\infty} \sqrt[n]{K} = 1$ with a simple limit comparison test but I can't figure out how exactly.

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 I think you meant (or should have meant) to write $$\lim_{n\to\infty}\sqrt [n] K=1\,$$ in your post's second line – DonAntonio Nov 5 '12 at 13:22 @DonAntonio you're right, fixed it. – hauptbenutzer Nov 5 '12 at 13:24

Hint: if $0<K<1$, $K=1/x$ for some $x>1$.
Hint: You can proove $\lim\limits_{n\to\infty}\sqrt[n]{n}=1$ by applying the Bernoulli inequality to $\sqrt{n}=(1+a_n)^{n/2}$ where $a_n=\sqrt{n}-1$.