Find the Second Term a_n=((n+2)!)/((n+3)!)

$a_{n} = \frac{(n + 2)!}{(n + 3)!}$

Substitute in the value of $n$ to find the $n$ th term.

$a_{2} = \frac{((2) + 2)!}{((2) + 3)!}$

Simplify the factorials.

Add $2$ and $2$ .

$a_{2} = \frac{(4)!}{((2) + 3)!}$

Add $2$ and $3$ .

$a_{2} = \frac{(4)!}{(5)!}$

Rewrite $(5)!$ as $5 \cdot 4!$ .

$a_{2} = \frac{(4)!}{5 \cdot 4!}$

Cancel the common factor of $(4)!$ .

Cancel the common factor.

$a_{2} = \frac{(4)!}{5 \cdot 4!}$

Rewrite the expression.

$a_{2} = \frac{1}{5}$