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

Consider $|a_n|=(1-\dfrac{3}{n})^n$ Now, $\lim\limits_{n \to \infty} |a_n|=\lim\limits_{n \to \infty} (1-\dfrac{3}{n})^n$ Thus, we have $=\lim\limits_{n \to \infty} (1+(-\dfrac{3}{n}))^n$ So, $\lim\limits_{n \to \infty} |a_n|=e^{-3}=\dfrac{1}{e^3}$ We need $\lim\limits_{n \to \infty} |a_n| =0$ for the series to converge, but we can see that $\lim\limits_{n \to \infty} |a_n| \ne 0$ Hence, The Series Diverges.