Answer
Converges to $e^{-1}$
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} [(\dfrac{1}{n})^{1/\ln n}]$
Since, $e^{\ln x}=x$
So, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} [(\dfrac{1}{n})^{1/\ln n}]= \lim\limits_{n \to \infty} e^{\frac{1}{\ln n} \ln (1/n)}= \lim\limits_{n \to \infty} e^{\frac{\ln 1 -\ln n}{\ln n}}=e^{-1}$
Hence, $\lim\limits_{n \to \infty} a_n=e^{-1}$ and {$a_n$} is Convergent and converges to $e^{-1}$
