$\lim_{n \to \infty} \frac{\ln(n)}{n} = 0$ proof with $\lim_{n \to \infty} n^{\frac{1}{n}} = 1$

Question

I have read, that you can proof $\lim_{n \to \infty} \frac{\ln(n)}{n} = 0$ with using $\lim_{n \to \infty} n^{\frac{1}{n}} = 1$

I know that $\frac{\ln(n)}{n}$ is equal to $n^{\frac{ln(ln(n))-ln(n)}{n}}$, but I don't know how I could use $\lim_{n \to \infty} n^{\frac{1}{n}} = 1$ to find the limit... Any hints?

$n^{1/n}=e^{(\ln n)/n}$. – André Nicolas Oct 23 '14 at 16:40 — André Nicolas, Oct 23 '14 at 16:40

score 2 · Answer 1 · answered Oct 23 '14 at 16:41

2

Hint

$$n^{\frac1n}=\exp\left(\frac{\ln n}{n}\right)$$

answered Oct 23 '14 at 16:41

score 1 · Answer 2 · answered Oct 23 '14 at 16:42

1

By continuity of $log$ function, $\lim_{n \to \infty} n^{\frac{1}{n}} = 1$ implies $\lim_{n \to \infty} \log n^{\frac{1}{n}} = \log 1=0$. Hence you get the result.

answered Oct 23 '14 at 16:42

John

13,204

score 0 · Answer 3 · answered Oct 23 '14 at 16:44

0

Ah, sorry...

Since $\lim_{n \to \infty} n^{\frac{1}{n}} = 1$ and $n^{\frac{1}{n}} = e^{\frac{\ln(n)}{n}}$, hence $\lim_{n \to \infty} \frac{\ln(n)}{n}$ must be $0$ because of $e^0 = 1$

answered Oct 23 '14 at 16:44

Rummelluff

489

$\lim_{n \to \infty} \frac{\ln(n)}{n} = 0$ proof with $\lim_{n \to \infty} n^{\frac{1}{n}} = 1$

3 Answers3

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