TITLE: How to solve this limit with factorial? $\lim_{n\to \infty}\frac{n!}{n^n}(\sum_{k=0}^n\frac{n^k}{k!}-\sum_{k=n+1}^\infty \frac{n^k}{k!})$
QUESTION [6 upvotes]: I want to solve the limit
$$\lim_{n\to \infty}\frac{n!}{n^n}\left(\sum_{k=0}^n\frac{n^k}{k!}-\sum_{k=n+1}^\infty \frac{n^k}{k!}\right)$$
This problem may be about Stirling's Approximation and Taylor series of exponent functions.
As $n$ is big enough, according to Stirling's approximation,
$$ n!\approx\sqrt{2\pi n}\frac{n^n}{e^n} $$
and according to Taylor series,
$$ \sum_{k=0}^\infty\frac{n^k}{k!}=e^n $$
The difference is the minus instead of the plus of the first $n$ term and the remainder.
I have tried to use the Taylor expansion with an integral remainder,
$$ e^n=\sum_{k=0}^n\frac{n^k}{k!}+\int_0^n\frac{(n-t)^n}{n!}e^t\mathrm{d}t $$
so we have the remainder as
$$ \sum_{k=n+1}^n\frac{n^k}{k!}=\int_0^n\frac{(n-t)^n}{n!}e^t\mathrm{d}t $$
and the first $n$ term as
$$ \sum_{k=0}^n\frac{n^k}{k!}=e^n-\int_0^n\frac{(n-t)^n}{n!}e^t\mathrm{d}t $$
Therefore, we have
\begin{aligned}
\frac{n!}{n^n}(\sum_{k=0}^n\frac{n^k}{k!}-\sum_{k=n+1}^\infty \frac{n^k}{k!})
&=\frac{n!}{n^n}\left(e^n-2\int_0^n\frac{(n-t)^n}{n!}e^t\mathrm{d}t\right)\\
&=\frac{n!e^n}{n^n}\left(1-2\int_0^n\frac{(n-t)^n}{n!}e^{-(n-t)} \mathrm{d}t\right)\\
&=\frac{n!e^n}{n^n}\left(1-\frac{2}{n!}\int_{0}^n s^ne^{-s}\mathrm{d}s\right)\\
&=\frac{n!e^n}{n^n}\left(1-\frac{2}{n!}\gamma(n+1,n)\right)\\
&\approx \sqrt{2\pi n}\left(1-\frac{2}{n!}\gamma(n+1,n)\right)\\
\end{aligned}
where the lower incomplete gamma function
$$ \gamma(n+1,n)=\int_{0}^n t^n e^{-t}\mathrm{d}t  $$
But how to continue?
And I also plot the scattering points of the series, we can see the limit is near 1.333 maybe.
How to continue the calculation and what is the limit?

REPLY [4 votes]: In this solution, we will prove that the limit is $4/3$ only using an elementary analysis. Let
$$ \bbox[background:#efffe0;padding:5px;]{S_n} = \frac{n!}{n^n} \left(\sum_{k=0}^{n}\frac{n^k}{k!} - \sum_{k=n+1}^{\infty} \frac{n^k}{k!}\right). $$
Then using OP's observation and the Stirling approximation,
\begin{align*}
S_n
&= \frac{n!}{n^n} \left( e^n - 2 \int_{0}^{n} \frac{(n-t)^n}{n!} e^{t} \, \mathrm{d}t \right) \\
&= \frac{n!}{n^n e^{-n}} - 2 \int_{0}^{n} \left(1 - \frac{t}{n}\right)^n e^{t} \, \mathrm{d}t \\
\quad &= \bigl( \sqrt{2\pi n} + o(1) \bigr) - 2 \sqrt{n} \bbox[background:#efffe0;padding:5px;]{ \underbrace{\int_{0}^{\sqrt{n}} \left(1 - \frac{s}{\sqrt{n}}\right)^n e^{\sqrt{n}s} \, \mathrm{d}s}_{=:I_n} }. \tag{$t = \sqrt{n}s$}
\end{align*}
Let's analyze the behavior of $I_n$. To this end, define $ \bbox[background:#efffe0;padding:5px;]{\phi_n(s)} = n\log\bigl(1-\frac{s}{\sqrt{n}}\bigr) + \sqrt{n}s $.
Then $I_n$ is recast as $I_n = \int_{0}^{\sqrt{n}} e^{\phi_n(s)} \, \mathrm{d}s$. Now we will make several observation.
$\text{(1)}$ Using the Maclaurin series of $\log(1-x)$, we find that $\phi_n(s) \leq -\frac{s^2}{2}$ holds for $0 \leq s \leq \sqrt{n}$. So, for any $0 < a \leq \sqrt{n}$, we have
$$
\int_{a}^{n} e^{\phi_n(s)} \, \mathrm{d}s
\leq \int_{a}^{n} e^{-s^2/2} \, \mathrm{d}s
\leq \int_{a}^{\infty} \frac{s}{a} e^{-s^2/2} \, \mathrm{d}s
= \frac{1}{a}e^{-s^2/2}. \tag{1}
$$
$\text{(2)}$ Define $ \bbox[background:#efffe0;padding:5px;]{\varepsilon_n(s)} := \phi_n(s) + \frac{s^2}{2} - \log\left(1-\frac{s^3}{3\sqrt{n}}\right) $. Using the Maclaurin series of $\log(1-x)$ again, for $1 < a < \frac{1}{2}n^{1/6}$ and $0 \leq s \leq a$, we have
$$
\left|\varepsilon_n(s)\right|
\leq n \sum_{k=4}^{\infty} \frac{s^k}{k n^{k/2}} + \sum_{l=2}^{\infty} \frac{s^{3l}}{3l n^{l/2}}
\leq \frac{Ca^6}{n}. \tag{2}
$$
for some absolute constant $C \in (0, \infty)$.
$\text{(3)}$ By integration by parts, we have
$$
\int_{a}^{\infty} s^3 e^{-s^2/2} \, \mathrm{d}s
= (a^2+2)e^{-a^2/2} \tag{3}
$$
To utilize these observations, we fix a sequence $(a_n)$ satisfying $\bbox[background:#efffe0;padding:5px;]{\sqrt{\log n} \ll a_n \ll n^{1/12}}$. (Here, $f(n) \ll g(n)$ means that $f(n) > 0$, $g(n) > 0$, and $f(n)/g(n) \to 0$ as $n\to\infty$.) For example, we can choose $(a_n)$ so that $a_n = \log n$ for large $n$. Then for sufficiently large $n$,
\begin{align*}
I_n
&= \int_{0}^{a_n} e^{\phi_n(s)} \, \mathrm{d}s + \mathcal{O}\biggl(\frac{e^{-a_n^2/2}}{a_n}\biggr) \tag{using (1)} \\
&= \int_{0}^{a_n} e^{-s^2/2}\left(1-\frac{s^3}{3\sqrt{n}}\right)e^{\varepsilon_n(s)} \, \mathrm{d}s + \mathcal{O}(a_n^{-1}e^{-a_n^2/2}) \tag{by def. of $\varepsilon_n$} \\
&= e^{\mathcal{O}(a_n^6/n)} \int_{0}^{a_n} e^{-s^2/2}\left(1-\frac{s^3}{3\sqrt{n}}\right) \, \mathrm{d}s + \mathcal{O}(a_n^{-1}e^{-a_n^2/2}) \tag{by (2)} \\
&= e^{\mathcal{O}(a_n^6/n)} \int_{0}^{\infty} e^{-s^2/2}\left(1-\frac{s^3}{3\sqrt{n}}\right) \, \mathrm{d}s + \mathcal{O}\bigl(a_n^2 e^{-a_n^2/2}\bigr) \tag{by (1) and (3)} \\
&= e^{\mathcal{O}(a_n^6/n)} \left( \sqrt{\frac{\pi}{2}} - \frac{2}{3\sqrt{n}} \right) + \mathcal{O}\bigl(a_n^2 e^{-a_n^2/2}\bigr)  \\
&= \sqrt{\frac{\pi}{2}} - \frac{2}{3\sqrt{n}} + \mathcal{O}(a_n^6/n) + \mathcal{O}\bigl(a_n^2 e^{-a_n^2/2}\bigr) 
\end{align*}
Plugging this back to $S_n$, it follows that
$$ S_n = \frac{4}{3} + \mathcal{O}(a_n^6/\sqrt{n}) + \mathcal{O}\bigl(\sqrt{n} a_n^2 e^{-a_n^2/2}\bigr) + o(1). $$
By the assumption on $(a_n)$, all the error terms vanish as $n \to \infty$, and therefore the desired claim follows.