Answer
First five terms: $1,-2,\dfrac{3}{2},\dfrac{-2}{3},\dfrac{5}{24}$
Work Step by Step
First five four of $a_n=\dfrac{(-1)^{n+1}n}{(n-1)!}$ are listed as below:
$a_1=\dfrac{(-1)^{1+1}}{(1-1)!}=1$;
$a_2=\dfrac{(-1)^{2+1}(2)}{(2-1)!}=-2$;
$a_3=\dfrac{(-1)^{3+1}(3)}{(3-1)!}=\dfrac{3}{2}$ ;
$a_4=\dfrac{(-1)^{4+1}(4)}{(4-1)!}=\dfrac{-4}{6}=\dfrac{-2}{3}$;
$a_5=\dfrac{(-1)^{5+1}(5)}{(5-1)!}=\dfrac{5}{24}$
First five terms: $1,-2,\dfrac{3}{2},\dfrac{-2}{3},\dfrac{5}{24}$
