Ratio Test — Question 2

*Question 2 Consider $\displaystyle\sum_{n=0}^{\infty}\frac{3^n}{n!}$.

1. Apply the Ratio Test and classify the series.

2. Recognize the resulting series and find its exact sum.

*Question 2 – Solution

With $a_n=3^n/n!$, $$\frac{a_{n+1}}{a_n}=\frac{3^{n+1}}{(n+1)!}\frac{n!}{3^n}=\frac3{n+1}.$$

$\displaystyle L=\lim_{n\to\infty}3/(n+1)=0<1$. The Ratio Test proves absolute convergence.

Since $\displaystyle e^x=\sum_{n=0}^{\infty}x^n/n!$, substituting $x=3$ gives $$\boxed{\displaystyle\sum_{n=0}^{\infty}\frac{3^n}{n!}=e^3}.$$