How to calculate the following sums?

Question

I would like to know of a way to evaluate the following two for arbitrary $n$.

$$\sum_{i=1}^ni!\,, \quad \sum_{i=1}^n \frac{n!}{i!}. $$

Answer 1

For the second sum: $$ \begin{align} \sum_{i=1}^n\frac{n!}{i!} &=\sum_{i=1}^\infty\frac{n!}{i!}-\sum_{i=n+1}^\infty\frac{n!}{i!}\\ &=(e-1)n!-\sum_{i=n+1}^\infty\frac{n!}{i!}\\ &=\lfloor(e-1)n!\rfloor \end{align} $$ Since $$ \begin{align} \sum\limits_{i=n+1}^\infty\frac{n!}{i!} &=\frac1{n+1}+\frac1{(n+1)(n+2)}+\frac1{(n+1)(n+2)(n+3)}+\dots\\ &\le\frac1{n+1}+\frac1{(n+1)^2}+\frac1{(n+1)^3}+\dots\\ &=\frac1n \end{align} $$

---

We can get an asymptotic expansion for the first sum: $$ \begin{align} \sum_{i=1}^ni! &=n!\left(1+\frac1n+\frac1{n(n-1)}+\frac1{n(n-1)(n-2)}+\dots\right)\\ &=n!\left(1+\frac1n+\frac1{n^2}+\frac2{n^3}+\frac5{n^4}+\frac{15}{n^5}+\dots\right) \end{align} $$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{k = 1}^{n}k!:\ {\large ?}.\qquad \sum_{k = 1}^{n}{n! \over k!}:\ {\large ?}}$

$\ds{\large\mbox{Attempt}\ {\tt I}}$: ( User $\tt @user2566092$ publishes this link about the first sum ). \begin{align} \sum_{k = 1}^{n}k!&=-1 + \sum_{k = 0}^{n}\int_{0}^{\infty}t^{k}\expo{-t}\,\dd t =-1 + \int_{0}^{\infty}\sum_{k = 0}^{n}t^{k}\expo{-t}\,\dd t =-1 + \int_{0}^{\infty}{t^{n + 1} - 1 \over t - 1}\,\expo{-t}\,\dd t \end{align}

$\ds{\large\mbox{Attempt}\ {\tt II}}$: \begin{align} \sum_{k = 1}^{n}k!&= \sum_{k = 0}^{n - 1}\Gamma\pars{k + 2} =\sum_{k = 0}^{\infty}\bracks{\Gamma\pars{k + 2} - \Gamma\pars{k + n + 2}} \\[3mm]&=\color{#c00000}{% \int_{0}^{\infty}\bracks{\Gamma\pars{x + 2} - \Gamma\pars{x + n + 2}}\,\dd x + \half\,\bracks{\Gamma\pars{2} - \Gamma\pars{n + 2}}} \\[3mm]&\color{#c00000}{\phantom{=}\mbox{}-2\Im\int_{0}^{\infty} {\Gamma\pars{\ic x + 2} - \Gamma\pars{\ic x + n + 2} \over \expo{2\pi x} - 1}\,\dd x} \\[3mm]&= \int_{0}^{\infty}\bracks{\Gamma\pars{x + 2} - \Gamma\pars{x + n + 2}}\,\dd x + \half - \half\,\pars{n + 1}! \\[3mm]&\phantom{=}\mbox{}-2\Im\int_{0}^{\infty} {\Gamma\pars{\ic x + 2} - \Gamma\pars{\ic x + n + 2} \over \expo{2\pi x} - 1}\,\dd x \end{align} $\color{#c00000}{\mbox{where we used the}}$ Abel-Plana Formula.

User $\tt @robjohn$ already solved the second one.

Answer 3

See http://mathworld.wolfram.com/FactorialSums.html for a rather complex formula for sum of factorials, albeit closed-form in terms of various functions.

Answer 4

For the first one you can have the integal representation

$$ \sum_{i=1}^{n} i! = \sum_{i=1}^{n} \Gamma(i+1) = \sum_{i=1}^{n} \int_{0}^{\infty}x^{i}e^{-x} dx =\int_{0}^{\infty} {\frac {{x}^{n+1}-x}{x-1}}e^{-x}dx $$