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

Question

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.

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 05 '12 at 13:22

score 3 · Accepted Answer · answered Nov 05 '12 at 13:19

3

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

answered Nov 05 '12 at 13:19

Javier

7,298

Brilliant, thank you. – hauptbenutzer Nov 05 '12 at 13:43

score 0 · Answer 2 · answered Nov 05 '12 at 13:29

0

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

answered Nov 05 '12 at 13:29

Christian Ivicevic

3,123

score 0 · Answer 3 · answered Dec 08 '13 at 19:33

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 $\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\sqrt[n]{K}=1$.

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

3 Answers3

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