Answer
$1,1,\dfrac{1}{2},\dfrac{1}{6}$
Work Step by Step
Need to find the first four terms of $a_n=\dfrac{1}{(n-1)!}$ when $n=1,2,3,4$
Thus,
$a_1=\dfrac{1}{(1-1)!}=1$;
$a_2=\dfrac{1}{(2-1)!}=1$ ;
$a_3=\dfrac{1}{(3-1)!}=\dfrac{1}{2}$ ;
and $a_4=\dfrac{1}{(4-1)!}=\dfrac{1}{6}$
Hence, the first four terms are: $1,1,\dfrac{1}{2},\dfrac{1}{6}$
