Answer: Option A is correct that is [tex]1 +1+\frac{1}{2} +\frac{1}{6}[/tex]
Explanation:
we will substitute the values of n in given expression
[tex]\sum_{n=1}^{4}\frac{n}{n!}[/tex]
when substituting n=1 we get in [tex]\sum_{n=1}^{4}\frac{n}{n!}[/tex]=[tex]\frac{1}{1!}[/tex]
when n=2 we get [tex]\frac{2}{2!}[/tex]
when n =3 we get [tex]\frac{3}{3!}=\frac{3}{6}=\frac{1}{2}[/tex] ;3 factorial that is 3! = 3 *2*1 = 6
when n=4 we get [tex]\frac{4}{4!}=\frac{4}{24}=\frac{1}{6}[/tex];4! = 4*3*2*1 = 24
Note: factorial means the product of the terms getting multiplied till 1
suppose n! will be equal to n(n-1)(n-2)(n-3).......1