# Question 822e7

Jan 26, 2018

The series converges.

#### Explanation:

Trying to find sum_(n=1)^oo(n^n/(n!*3^n)) (I'm staring at $n = 1$ instead of $n = 0$ because of the ${0}^{0}$ when $n = 0$.)

Using the Ratio Test:
lim_(n to oo)|(n+1)^(n+1)/((n+1)!\*3^(n+1))*(n!\*3^n)/n^n|

=lim_(n to oo)|((n+1)(n+1)^(n))/((n+1)\*n!\*3^(n)\*3)*(n!\*3^n)/n^n|#

$= {\lim}_{n \to \infty} | {\left(n + 1\right)}^{n} / \left(3 {n}^{n}\right) |$

$= \frac{1}{3} {\lim}_{n \to \infty} | {\left(\frac{n + 1}{n}\right)}^{n} |$

$= \frac{1}{3} {\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{n}$

$= \frac{1}{3} \cdot e$ because ${\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{n}$ is famously equal to $e$. See this video for the work.

Since $\frac{e}{3} < 1$, the series converges.