## Calculus: Early Transcendentals 8th Edition

$e^{3}-1$
Given: $3+\frac{9}{2!}+\frac{27}{3!}+....=\Sigma_{n=1}^{\infty}\dfrac{3^{n}}{(n!)}$ The above series resembles: $e^{x}-1=\Sigma_{n=1}^{\infty}\dfrac{x^{n}}{(n!)}$ Thus, $\Sigma_{n=1}^{\infty}\dfrac{3^{n}}{(n!)}=e^{3}-1$