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