Answer
Converges to $e^{-1}$
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (1-\dfrac{1}{n})^{n}$
Since, $\lim\limits_{n \to \infty} (1+\dfrac{x}{n})^{n}=e^x$ when $x \gt 0$
So, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (1-\dfrac{1}{n})^{n}=\lim\limits_{n \to \infty} (1+(-\dfrac{1}{n}))^{n}=e^{-1}$
Hence, $\lim\limits_{n \to \infty} a_n=e^{-1}$ and {$a_n$} is convergent and converges to $e^{-1}$
