Answer
converges to $e^{-1}$
Work Step by Step
As we know that when $x \gt 0$, $\lim\limits_{n \to \infty} (1+\dfrac{x}{n})^{n}=e^x$
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (\dfrac{n}{n+1})^{n}$
This implies that $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (\dfrac{n}{n+1})^{n}$
and $a_n= \lim\limits_{n \to \infty} (1+\dfrac{1}{-(n+1)})^{n}=e^{ \lim\limits_{n \to \infty}[(\dfrac{n}{-(n+1)})]}=e^{-1}$
Hence, $\lim\limits_{n \to \infty} a_n=e^{-1}$ and {$a_n$} converges to $e^{-1}$
