## Thomas' Calculus 13th Edition

converges to $e^{-1}$
As we know that when $x \gt 0$ so, $\lim\limits_{n \to \infty} \sqrt[n] {n}=1$ and $\lim\limits_{n \to \infty} x^{1/n}=1$ and $e^{\ln x}=x$ Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} [(\dfrac{1}{n})^{1/\ln n}]$ This implies that $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} [(\dfrac{1}{n})^{1/\ln n}]$ and $a_n= \lim\limits_{n \to \infty} (e)^{\frac{1}{\ln n} \ln (1/n)}=\lim\limits_{n \to \infty} e^{\frac{\ln 1 -\ln n}{\ln n}}=e^{-1}$ Thus, $\lim\limits_{n \to \infty} a_n=e^{-1}$ and {$a_n$} converges to $e^{-1}$.