I want to estimate $f\_p(e/p)$ as $p\to \infty$ for the series 
$$
f\_p(x) = \sum\_{k=1}^p k! x^k
$$


I am sure this question in the right context is easy but I haven't found the information on how to compute it.



Asymptotic Estimation of the Factorial Power Series
Denote $k=p-m$. Then $$k!x^k=(p!x^p)e^{-m}\cdot\frac{p^m}{p(p-1)\dots (p-m+1)}.$$
For any fixed $m$ the fractions tends to 1. It follows that $$\lim \frac{\sum\_{m=0}^N (p-m)!x^{p-m}}{p! x^p}=\sum\_{m=0}^N e^{-m},$$
hence $$\liminf \frac{\sum\_{m=0}^{p-1} (p-m)!x^{p-m}}{p! x^p}\geq \sum\_{m=0}^\infty e^{-m}=\frac{e}{e-1}.$$
I claim that this liminf is a genuine limit. Now it suffices to get upper estimate of the form $$\frac{\sum\_{m=N}^{p-1} (p-m)!x^{p-m}}{p! x^p}\leq \varepsilon\_N$$
for large enough $p$, where $\varepsilon\_N$ tends to 0 when $N$ grows. We use Stirling approximation for factorial: upto absolute multiplicative constant we may replace each $k!$ to $\sqrt{k}(k/e)^k$. Then, since $\sqrt{p-m}\leq \sqrt{p}$ for each $m$, it suffices to get an estimate of the form $$\sum\_{m=N}^{p-1} \left(\frac{p-m}p\right)^{p-m}\leq \varepsilon\_N.$$
Denote $p-m=tp$, $0<t<1$, then $F(m):=(1-m/p)^{p-m}=t^{tp}$. Function $t^t$ decreases upto $1/e$ and increases after $1/e$. It means that $F(p-1)=1/p$, $F(p-2)=4/p^2\geq F(p-s)$ for all $s\leq p/e$. It follows that $\sum\_{s\leq p/e} F(p-s)\leq 5/p$. This is ok. For $p-m>p/e$ we have 
$$
F(m)=(1-m/p)^{p-m}\leq (1-m/p)^{p/e}\leq e^{-m/e},
$$
and we may define $\varepsilon\_N=2\sum\_{m\geq N} e^{-m/e}$.


