Just wondering if somebody can help out with the workings for the following:- lim n-> inf (ln n)^1/n
$\displaystyle (\ln{n})^{\frac{1}{n}} = e^{\ln{[(\ln{n})^{\frac{1}{n}}]}}$
$\displaystyle = e^{\frac{1}{n}\ln{(\ln{n})}}$
$\displaystyle = e^{\frac{\ln{(\ln{n})}}{n}}$.
Therefore:
$\displaystyle \lim_{n \to \infty}(\ln{n})^{\frac{1}{n}} = \lim_{n \to \infty} e^{\frac{\ln{(\ln{n})}}{n}}$
$\displaystyle = e^{\lim_{n \to \infty}\frac{\ln{(\ln{n})}}{n}}$
$\displaystyle = e^{\lim_{n \to \infty}\frac{\frac{1}{n\ln{n}}}{1}}$ by L'Hospital's Rule
$\displaystyle = e^{\lim_{n \to \infty}\frac{1}{n\ln{n}}}$
$\displaystyle = e^0$
$\displaystyle = 1$.