Answer
$a_{n}=\displaystyle \frac{1}{n!}$
Work Step by Step
$\displaystyle \frac{1}{2}=1\cdot\frac{1}{2}$ (second term = previous$\displaystyle \times\frac{1}{2})$
$\displaystyle \frac{1}{6}=\frac{1}{3}\cdot\frac{1}{2}$ (third term = previous$\displaystyle \times\frac{1}{3})$
$\displaystyle \frac{1}{24}=\frac{1}{6}\cdot\frac{1}{4}$ ( fourth term = previous$\displaystyle \times\frac{1}{3})$
fifth term = fourth$\displaystyle \times\frac{1}{5}$
So, with this pattern,
$a_{1}=1$
$a_{2}=\displaystyle \frac{a_{1}}{2}$
$a_{3}=\displaystyle \frac{a_{1}}{2\cdot 3}$
$a_{4}=\displaystyle \frac{a_{1}}{2\cdot 3\cdot 4}=\frac{a_{1}}{4!}=\frac{1}{4!}$
$a_{5}=\displaystyle \frac{1}{4!\cdot 5}=\frac{1}{5!}$
$...$
$a_{n}=\displaystyle \frac{1}{n!}$
