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

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    $\begingroup$ A generating function for this sum is given by $$S = [z^n] (e^z-1-z) / (1-z)$$. This does not involve the gamma function. $\endgroup$
    – mds
    May 24, 2018 at 22:26

2 Answers 2

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

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  • $\begingroup$ Just to note a minor extension that I obtained recently following this prescription: $\sum_{k=0}^n\frac{m^{n-k}n!}{k!} = \lfloor m^n n! e^{\frac{1}{m}}\rfloor$ $\endgroup$
    – Roger V.
    Sep 30, 2020 at 9:08
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Let's rewrite this as $\;\displaystyle S=\sum _{j=0}^{n} \frac{1}{j!}-2\;$ then prove that $$\sum _{j=0}^{n} \frac{1}{j!}=\frac{\lfloor e\; n!\rfloor}{n!}$$

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  • $\begingroup$ Can you explain it a bit. Is $$\sum _{j=0}^{n} \frac{1}{j!}=\frac{\lfloor e\; n!\rfloor}{n!}$$ a standard result? $\endgroup$
    – me_ravi_
    May 14, 2015 at 8:35
  • $\begingroup$ wonderful... any idea how to prove? $\endgroup$ May 14, 2015 at 8:35
  • $\begingroup$ @user3127040: To prove it consider the remainder of the exponential expansion and multiply it by $n!$ you should obtain something like $\;\displaystyle \frac 1{(n+1)}+ \frac 1{(n+1)(n+2)} +\cdots$ that you may bound with a geometric series smaller than $1$. $\endgroup$ May 14, 2015 at 8:41
  • $\begingroup$ @Integrator: Yes a standard result perhaps from the irrationality of $e$ (possibly earlier I don't know...). $\endgroup$ May 14, 2015 at 8:43
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    $\begingroup$ @RaymondManzoni: Thanks a lot. $\endgroup$
    – me_ravi_
    May 14, 2015 at 8:56

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