Answer
Diverges and $\lim\limits_{n \to \infty} a_n=\dfrac{1}{e^2}$
Work Step by Step
Let us consider $a_n=(\dfrac{n-2}{n})^n$
In order to solve the given series we will take the help of Ratio Test. This test states that when the limit $L \lt 1$, the series converges and for $L \gt 1$, the series diverges.
Then $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} (\dfrac{n-2}{n})^n=\lim\limits_{n \to \infty} (1-\dfrac{2}{n})$
$\implies L=\lim\limits_{n \to \infty} a_n=e^{-2}=\dfrac{1}{e^2}$
Thus, $\lim\limits_{n \to \infty} a_n \ne 0$
Hence, the Series Diverges and $\lim\limits_{n \to \infty} a_n=\dfrac{1}{e^2}$
