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

Consider $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \sinh (\ln n)$ Since, $e^{\ln x}=x$ So, $\lim\limits_{n \to \infty} \sinh (\ln n)=\lim\limits_{n \to \infty} \dfrac{e^{\ln n}-e^{-\ln n}}{2}=\lim\limits_{n \to \infty} \dfrac{n-1/n}{2}$ or, $=\infty$ Hence, $\lim\limits_{n \to \infty} a_n=\infty$ and {$a_n$} is Divergent