Answer
$1,2,\dfrac{3}{2},\dfrac{2}{3}$
Work Step by Step
Need to find the first four terms of $a_n=\dfrac{n^2}{n!}$ when $n=1,2,3,4$
Thus,
$a_1=\dfrac{1^2}{1!}=1$
$a_2=\dfrac{2^2}{2!}=2$
$a_3=\dfrac{3^2}{3!}=\dfrac{9}{3 \cdot 2 \cdot 1}=\dfrac{3}{2}$
$a_4=\dfrac{4^2}{4!}=\dfrac{16}{4 \cdot 3 \cdot 2 \cdot 1}=\dfrac{2}{3}$
Hence, the first four terms are: $1,2,\dfrac{3}{2},\dfrac{2}{3}$
