## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{(2(n+1))!}{((n+1)!)^{2}}}{\frac{(2n)!}{(n!)^{2}}}|$ $=\lim\limits_{n \to \infty}\frac{2(n+1))(2n+1)}{(n+1)^{2}}$ $=\lim\limits_{n \to \infty}\frac{2(2n+1))}{(n+1)}$ $=\lim\limits_{n \to \infty}\frac{4+2/n}{1+1/n}$ $=4 \gt 1$ The series is divergent.