Ratio Test — Question 2

Question 2 – Solution

Step 1: Form the ratio.

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

Step 2: Take the limit.

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

Step 3: Identify the sum.

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