Finite Summation of Fractional Factorial Series

Question

Is there a closed form solution for the following series? (Without Using Gamma Function): $$ S=\sum _{i=1}^{n-1} \frac{1}{(i+1)!} $$

Answer 1

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 $$\lfloor n!\,e\rfloor=\sum_{k=0}^n\frac{n!}{k!}\\ \implies \frac{\lfloor n!\,e\rfloor}{n!}=\sum_{k=0}^n\frac{1}{k!}\\ \implies \frac{\lfloor n!\,e\rfloor}{n!}-2=\sum _{k=1}^{n-1} \frac{1}{(k+1)!} $$

Answer 2

Let's rewrite this as $\;\displaystyle S=\sum _{j=0}^{n} \frac{1}{j!}-2\;$ then prove that $$\sum _{j=0}^{n} \frac{1}{j!}=\frac{\lfloor e\; n!\rfloor}{n!}$$