Question

4Σn=! n/n!

Answer 1

I think the sum is supposed to be

$\displaystyle\sum_{n=1}^4\frac n{n!}=\sum_{n=1}^4\frac1{(n-1)!}$

since $n!=n\cdot(n-1)!$. Then

$\displaystyle\sum_{n=1}^4\frac1{(n-1)!}=\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}$

and $0!=1$ by definition so that the sum has a value of $\dfrac83$.