Answer
$1,-2,\dfrac{3}{2},\dfrac{-2}{3},\dfrac{5}{24}$
Work Step by Step
Need to find the first four terms of $a_n=\dfrac{(-1)^{n+1}n}{(n-1)!}$ when $n=1,2,3,4$
Thus,
$a_1=\dfrac{(-1)^{1+1}}{(1-1)!}=1$
$a_2=\dfrac{(-1)^{2+1}(2)}{(2-1)!}=-2$
$a_3=\dfrac{(-1)^{3+1}(3)}{(3-1)!}=\dfrac{3}{2}$
$a_4=\dfrac{(-1)^{4+1}(4)}{(4-1)!}=\dfrac{-2}{3}$
$a_5=\dfrac{(-1)^{5+1}(5)}{(5-1)!}=\dfrac{5}{24}$
Hence, the first five terms are: $1,-2,\dfrac{3}{2},\dfrac{-2}{3},\dfrac{5}{24}$
