## University Calculus: Early Transcendentals (3rd Edition)

Diverges and $\lim\limits_{n \to \infty} a_n=\dfrac{1}{e^2}$
Consider $a_n=(\dfrac{n-2}{n})^n$ Now, $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} (\dfrac{n-2}{n})^n$ Thus, we have $\lim\limits_{n \to \infty}(\dfrac{n}{n}-\dfrac{2}{n})=\lim\limits_{n \to \infty} (1-\dfrac{2}{n})$ So, $\lim\limits_{n \to \infty} a_n=e^{-2}=\dfrac{1}{e^2}$ Thus, $\lim\limits_{n \to \infty} a_n =0$ for the series to converge, but we can see that $\lim\limits_{n \to \infty} a_n \ne 0$ Hence, The Series Diverges and $\lim\limits_{n \to \infty} a_n=\dfrac{1}{e^2}$