TITLE: Finite Summation of Fractional Factorial Series
QUESTION [6 upvotes]: Is there a closed form solution for the following series? (Without Using Gamma Function):
$$
S=\sum _{i=1}^{n-1} \frac{1}{(i+1)!}
$$

REPLY [14 votes]: Note that
$$n!\,e=\sum_{k=0}^\infty\frac{n!}{k!}=\sum_{k=0}^n\frac{n!}{k!}+\sum_{k=n+1}^\infty\frac{n!}{k!}$$
The first sum on the RHS is always an integer since $n\geq k$. The second sum satisfies
$$\begin{align} \sum_{k=n+1}^\infty\frac{n!}{k!}
&=\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots\\
&<\frac{1}{n+1}+\frac{1}{(n+1)(n+1)}+\frac{1}{(n+1) (n+1)(n+1)}+\cdots\\
&=\sum_{k=1}^\infty\frac{1}{(n+1)^k}\\
&= \frac{1}{n}\\
&\leq1
\end{align}$$
when $n\geq1$. Hence we have
$$\lfloor n!\,e\rfloor=\sum_{k=0}^n\frac{n!}{k!}\\
\implies \frac{\lfloor n!\,e\rfloor}{n!}=\sum_{k=0}^n\frac{1}{k!}\\
\implies \frac{\lfloor n!\,e\rfloor}{n!}-2=\sum _{k=1}^{n-1} \frac{1}{(k+1)!}
$$