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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
up vote 3 down vote accepted

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

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+1 As simple as that. – DonAntonio Nov 5 '12 at 13:22
Brilliant, thank you. – hauptbenutzer Nov 5 '12 at 13:43

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 $a_n=\sqrt{n}-1$.

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From the root test it follows that for a positive sequence $(a_n)_{n\in\mathbb N}$, if $\lim_{n\to\infty}\sqrt[n]{a_n}<1$ then $\lim_{n\to\infty}a_n=0$ while if $\lim_{n\to\infty}\sqrt[n]{a_n}>1$ then $\lim_{n\to\infty}a_n=+\infty$.
Since for the sequence $a_n=K$, we have $\lim_{k\to\infty} a_n=K>0$ it follows that $\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\sqrt[n]{K}=1$.

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