Limit proof

**tttcomrader** · October 3rd 2008, 05:24 AM

Prove that $\lim _{n \rightarrow \infty } n^ { \frac {1}{n} } = 1$

Proof so far.

Define the sequence $\{ a_n \} = n^ { \frac {1}{n} } -1$

I want to show that $\{ a_n \} \rightarrow 0$

**CaptainBlack** · October 3rd 2008, 07:28 AM

Originally Posted by tttcomrader

Prove that $\lim _{n \rightarrow \infty } n^ { \frac {1}{n} } = 1$

Proof so far.

Define the sequence $\{ a_n \} = n^ { \frac {1}{n} } -1$

I want to show that $\{ a_n \} \rightarrow 0$

Take logs and show that

$\lim _{n \to \infty } \frac {\ln(n)}{n} = 0$

You can do this by switching to a continuous variable and then using L'Hopital's rule

RonL

**ThePerfectHacker** · October 3rd 2008, 10:42 AM

Originally Posted by tttcomrader

Prove that $\lim _{n \rightarrow \infty } n^ { \frac {1}{n} } = 1$

Proof so far.

Define the sequence $\{ a_n \} = n^ { \frac {1}{n} } -1$

I want to show that $\{ a_n \} \rightarrow 0$

First $a_n \geq 0$ . Second $(a_n + 1)^n = n$ .
Therefore, $n = (a_n+1)^n \geq 1 + na_n + \tfrac{1}{2}n(n-1)a_n^2 \geq \tfrac{1}{2}n(n-1)a_n^2$

Thus, $0 \leq a_n \leq \sqrt{\frac{2}{n-1}}$

Now use squeeze theorem to show $\lim ~ a_n = 0$ .

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