Ratio Test — Question 1

*Question 1 Consider $\displaystyle\sum_{n=1}^{\infty}\frac{n!}{3^n}$.

1. Compute $a_{n+1}/a_n$.

2. Decide whether the series converges.

3. Explain what the calculation says about factorial growth compared with $3^n$.

*Question 1 – Solution

Let $a_n=n!/3^n$. Its terms are positive, so absolute values do not change the ratio.

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

Thus $\displaystyle L=\lim_{n\to\infty}(n+1)/3=\infty>1$. The terms eventually increase; indeed, $a_{n+1}>a_n$ for $n>2$. Therefore $a_n$ cannot approach $0$.

The series diverges by the Ratio Test (and by the nth-term test). The unbounded ratio shows that $n!$ eventually grows faster than $3^n$.