## Calculus: Early Transcendentals 8th Edition

$\Sigma_{n=1}^{ \infty}\frac{2.4.6.........(2n)}{n!}=\Sigma_{n=1}^{ \infty}\frac{2^{n}n!}{n!}$ Here, $\Sigma_{n=1}^{ \infty}2^{n}$ is a geometric series with common ratio $r= 2\geq 1$, thus the series diverges.