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

real-analysis sequences-and-series convergence-divergence calculus limits

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.


Hint: if $0<K<1$, $K=1/x$ for some $x>1$.

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