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