Asymptotic behavior of tail series $A_n=\sum_{m\ge n+1} \frac{n!}{m!}$ invoked by $e$ Recalling a series, denote
$$
A_n=\sum_{m=n+1}^{\infty} {\frac{n!}{m!}}
$$
which is relative to the tail of $e$, and sometime emerges in some well-known limits like
$$
\lim_{n\to\infty} n\sin(2\pi e n!) = 2\pi
$$
as $n\to\infty$, the very first order of $A_n$ is trivial, since
$$
\frac1{n+1}<A_n<\frac1{n-1}
$$
so
$$
A_n \sim \frac1{n} + o(n^{-1}) 
$$
when it comes to the higher order, a possible method I used is using the Euler–Maclaurin formula with reciprocal gamma function, which may be an arguably inconvenient approach for this 'simple' series. Here, may I ask for other quick methods to its asymptotic form, some first items of which I found is (may not correct)
$$
A_n \sim \frac1{n} - \frac1{n^3} + \frac1{n^4} + o(n^{-4}) 
$$
where the squared item is happened to be missing.
 A: We can write your sum as
$$
A_n=n!e-\sum_{m=0}^n \frac {n!}{m!}=n!e-s_n
$$
now we use $r!=\int_{R_+}t^{r}e^{-t}$ and $\binom{r}{p}=\frac{r!}{(r-p)!p!}$
to get
$$
s_n=\int_{R_+}e^{-t}t^{n}\sum_{m=0}^n\binom{n}{m}t^{-m}=\int_{R_+}e^{-t}(1+t)^n
$$
by shifting $t \rightarrow l-1$ we get
$$
s_n=e\int_{1}^{\infty}e^{-l}l^n=n!e-e\int_{0}^{1}e^{-l}l^n=n!e-eI_n
$$
Now the asymptotics of $I_n$ can now be obtained to arbitrary order using repeated integration by parts.
For example to second order:
$$
eI_n=\frac{1}{n}-\frac{1}{n^3}+o(n^{-3})
$$
or
$$
A_n = \frac{1}{n}-\frac{1}{n^3}+o(n^{-3})
$$
A: Using the beta integral, we have
\begin{align*}
A_n  &= \sum\limits_{k = 1}^\infty  {\frac{1}{{(n + 1)(n + 2) \cdots (n + k)}}}  = \sum\limits_{k = 1}^\infty  {\frac{1}{{(k - 1)!}}\int_0^1 {t^n (1 - t)^{k - 1} dt} } \\ & = \int_0^1 {t^n e^{1 - t} dt}  = \int_0^{ + \infty } {e^{ - ns} e^{1 - s - e^{ - s} } ds} 
\end{align*}
for all $n\geq 1$. We have
$$
e^{1 - s - e^{ - s} }  = 1 - \frac{{s^2 }}{2} + \frac{{s^3 }}{6} + \frac{{s^4 }}{{12}} -  \cdots 
$$
near $s=0$. Thus, by Watson's lemma,
$$
A_n  \sim \frac{1}{n} - \frac{1}{{n^3 }} + \frac{1}{{n^4 }} + \frac{2}{{n^5 }} -  \cdots 
$$
as $n\to +\infty$. It is possible to express the coefficients of this asymptotic expansion in terms of Stirling numbers of the second kind.
A: We can rewrite $A_n$ in multiple ways
$$
\eqalign{
  & A_{\,n}  = \sum\limits_{n + 1\, \le \,m} {{{n!} \over {m!}}}  = \sum\limits_{0\, \le \,k}
 {{{n!} \over {\left( {n + 1 + k} \right)!}}}  =   \cr 
  &  = \sum\limits_{0\, \le \,k} {{1 \over {\left( {n + 1} \right)^{\,\overline {\,k + 1\,} } }}}
  = \sum\limits_{0\, \le \,k} {n^{\,\underline {\, - \left( {k + 1} \right)\,} } }  =   \cr 
  &  = {1 \over {n + 1}}\sum\limits_{0\, \le \,k} {{1 \over {\left( {n + 2} \right)^{\,\overline {\,k\,} } }}}
  = {1 \over {n + 1}}{}_1F_{\,1} \left( {\left. {\matrix{  1  \cr  {n + 2}  \cr  } \;} \right|\;1} \right) =   \cr 
  &  = \Gamma \left( {n + 1} \right)\sum\limits_{0\, \le \,k} {{1 \over {\Gamma \left( {n + 1 + k + 1} \right)}}}
  = \;e\;\gamma \left( {n + 1,1} \right) \cr} 
$$
where:

*

*$n^{\,\underline {\,k\,} } ,\quad n^{\,\overline {\,k\,} } $ represent respectively the
Falling and Rising Factorial;

*${}_1F_{\,1}$ is the Confluent Hypergeometric Function;

*$\gamma(s,z)$ is the Lower Incomplete Gamma Function.

From the expression in the Rising Factorial, inverting $n$ into $1/z$ we get the asymptotics
$$
\begin{array}{l}
 \frac{1}{{\left( {n + 1} \right)^{\,\overline {\,k + 1\,} } }}\quad \left| {\;z = } \right.\frac{1}{n}\quad  = \frac{1}{{\left( {\frac{1}{z} + 1} \right)^{\,\overline {\,k + 1\,} } }} =  \\ 
  = \frac{1}{{\left( {\frac{1}{z} + 1} \right)\left( {\frac{1}{z} + 2} \right) \cdots \left( {\frac{1}{z} + k + 1} \right)}} =  \\ 
  = \frac{{z^{\,\left( {k + 1} \right)} }}{{\left( {z + 1} \right)\left( {2\,z + 1} \right) \cdots \left( {\left( {k + 1} \right)\,z + 1} \right)}}\quad \left| {\,\left| {\,z\,} \right| < \frac{1}{{k + 1}}} \right.\quad  =  \\ 
  = z^{\,k + 1} \left( {\sum\limits_{0\, \le \,l_{\,1} \,} {\left( { - z} \right)^{\,\,l_{\,1} } } } \right)\left( {\sum\limits_{0\, \le \,\,\,l_{\,2} \,} {\left( { - 2\,z} \right)^{\,\,l_{\,2} } } } \right) \cdots \left( {\sum\limits_{0\, \le \,\,\,l_{\,k + 1} \,} {\left( { - \left( {k + 1} \right)\,z} \right)^{\,\,l_{\,k + 1} } } } \right) =  \\ 
  = z^{\,k + 1} \sum\limits_{0\, \le \,s\,} {\left( { - 1} \right)^{\,\,s} \left( {\sum\limits_{\scriptstyle \left\{ {\begin{array}{*{20}c}
   {0\, \le \,l_{\,j} }  \\
   {l_{\,1}  + l_{\,2}  +  \cdots l_{\,k + 1}  = s}  \\
\end{array}} \right.  \atop 
  \scriptstyle \, } {\prod\limits_{1\, \le \,j\, \le \,k + 1} {j^{\,\,l_{\,j} } } } } \right)\;z^{\,\,s} }  =  \\ 
  = z^{\,\,k + 1} \sum\limits_{0\, \le \,s\,} {\left( { - 1} \right)^{\,\,s} \left\{ \begin{array}{c}
 s + \,k + 1 \\ 
 \,k + 1 \\ 
 \end{array} \right\}\;z^{\,\,s} }  = \sum\limits_{0\, \le \,s\,} {\left( { - 1} \right)^{\,\,s - \,k - 1} \left\{ \begin{array}{c}
 s \\ 
 \,k + 1 \\ 
 \end{array} \right\}\;z^{\,\,s} }  \\ 
 \end{array}
$$
or more simply
$$
\eqalign{
  & {1 \over {\left( {n + 1} \right)^{\,\overline {\,k + 1\,} } }} = n^{\,\underline {\, - \left( {k + 1} \right)\,} }  =   \cr 
  &  = \sum\limits_{0\, \le s\,} {\left( { - 1} \right)^{\,\,\, - \left( {k + 1} \right) - s}
 \left[ \matrix{ - \left( {k + 1} \right) \cr   s \cr}  \right]\;n^{\,\,s} }  =   \cr 
  &  = \sum\limits_{0\, \le s\,} {\left( { - 1} \right)^{\,\,\,k + 1 - s}
 \left\{ \matrix{ - s \cr   k + 1 \cr}  \right\}\;n^{\,\,s} }  =   \cr 
  &  = \sum\limits_{0\, \le s\,} {\left( { - 1} \right)^{\,\,\,k + 1 - s}
 \left\{ \matrix{ s \cr   k + 1 \cr}  \right\}\;n^{\,\, - \,s} }  \cr} 
$$
Thus
$$
\eqalign{
  & A_{\,n}  = \sum\limits_{0\, \le s\,} {\left( {\sum\limits_{0\, \le \,k\,\left( { \le \,s - 1} \right)}
 {\left( { - 1} \right)^{\,\,\,k + 1 - s} \left\{ \matrix{  s \cr  k + 1 \cr}  \right\}\;} } \right)n^{\,\, - \,s} }  =   \cr 
  &  = \sum\limits_{0\, \le s\,} {{{c_{\,s} } \over {n^{\,s} }}}   \cr 
  & c_{\,s}  = 0,1,0, - 1, - 1,2, - 9,9,50, - 267,413,2180, \cdots  \cr} 
$$
But the Stirling Numbers are known to grow very rapidly.
Here in fact is a lin-log graph of the coefficients.

The expression in Incomplete Gamma instead gives
$$
\eqalign{
  & A_{\,n}  = \;e\;\gamma \left( {n + 1,1} \right) \sim   \cr 
  &  \sim {{\Gamma \left( {n + 1} \right)e^{\,n + 1} } \over {\sqrt {2\pi } \left( {n + 1} \right)^{\,n + 3/2} }}
\left( {1 + {{11} \over {12\left( {n + 1} \right)}} - {{23} \over {288\left( {n + 1} \right)^{\,2} }}
 + O\left( {{1 \over {n^{\,3} }}} \right)} \right) \cr} 
$$
