# Limit proof

• Oct 3rd 2008, 05:24 AM
Limit proof
Prove that $\displaystyle \lim _{n \rightarrow \infty } n^ { \frac {1}{n} } = 1$

Proof so far.

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

I want to show that $\displaystyle \{ a_n \} \rightarrow 0$
• Oct 3rd 2008, 07:28 AM
CaptainBlack
Quote:

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

Proof so far.

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

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

Take logs and show that

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

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

RonL
• Oct 3rd 2008, 10:42 AM
ThePerfectHacker
Quote:

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

Proof so far.

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

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

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

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

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