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

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

 A: 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:
$$
\begin{align}
\sum_{i=1}^ni!
&=n!\left(1+\frac1n+\frac1{n(n-1)}+\frac1{n(n-1)(n-2)}+\dots\right)\\
&=n!\left(1+\frac1n+\frac1{n^2}+\frac2{n^3}+\frac5{n^4}+\frac{15}{n^5}+\dots\right)
\end{align}
$$
A: For the first one you can have the integal representation

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

A: See http://mathworld.wolfram.com/FactorialSums.html for a rather complex formula for sum of factorials, albeit closed-form in terms of various functions.
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\sum_{k = 1}^{n}k!:\ {\large ?}.\qquad
     \sum_{k = 1}^{n}{n! \over k!}:\ {\large ?}}$

$\ds{\large\mbox{Attempt}\ {\tt I}}$: ( 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.

