I think the sum is supposed to be
[tex]\displaystyle\sum_{n=1}^4\frac n{n!}=\sum_{n=1}^4\frac1{(n-1)!}[/tex]
since [tex]n!=n\cdot(n-1)![/tex]. Then
[tex]\displaystyle\sum_{n=1}^4\frac1{(n-1)!}=\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}[/tex]
and [tex]0!=1[/tex] by definition so that the sum has a value of [tex]\dfrac83[/tex].
