Evaluating $\lim_{n\to \infty}\frac{n!}{n^n}\left(\sum_{k=0}^{n}\frac{n^k}{k!}-\sum_{k=n+1}^{\infty}\frac{n^k}{k!}\right)$ How do you evaluate the following limit:
$$\lim_{n\to \infty}\frac{n!}{n^n}\left(\sum_{k=0}^{n}\frac{n^k}{k!}-\sum_{k=n+1}^{\infty}\frac{n^k}{k!}\right)$$
 A: Here is a partial solution towards the following claim:

Claim. We have
  $$ A_n
:= \frac{n!}{n^n} \left( \sum_{k=0}^{n} \frac{n^k}{k!} - \sum_{k=n+1}^{\infty} \frac{n^k}{k!} \right)
\xrightarrow[n\to\infty]{} \frac{4}{3}. $$

This claim is supported by numerical experiments, and we aim at justifying this claim.

Step 1. Notice that
$$ \sum_{k=0}^{n} \frac{n^k}{k!}
= e^n \int_{n}^{\infty} \frac{x^n e^{-x}}{n!} \, \mathrm{d}x
\stackrel{(x=n+\sqrt{n} u)}=\frac{n^{n+\frac{1}{2}}}{n!} \int_{0}^{\infty} \left(1 + \frac{u}{\sqrt{n}}\right)^{n}e^{-\sqrt{n}u} \, \mathrm{d}u $$
and similarly
$$ \sum_{k=n+1}^{\infty} \frac{n^k}{k!}
= e^n \int_{0}^{n} \frac{x^{n}e^{-x}}{n!} \, \mathrm{d}x
\stackrel{(x=n-\sqrt{n} u)}=\frac{n^{n+\frac{1}{2}}}{n!} \int_{0}^{\sqrt{n}} \left(1 - \frac{u}{\sqrt{n}}\right)^{n}e^{\sqrt{n}u} \, \mathrm{d}u. $$
Combining altogether, we get
\begin{align*}
A_n
&:= \frac{n!}{n^n} \left( \sum_{k=0}^{n} \frac{n^k}{k!} - \sum_{k=n+1}^{\infty} \frac{n^k}{k!} \right) \\
&= \sqrt{n} \left( \int_{0}^{\infty} \left(1 + \frac{u}{\sqrt{n}}\right)^{n}e^{-\sqrt{n}u} \, \mathrm{d}u - \int_{0}^{\sqrt{n}} \left(1 - \frac{u}{\sqrt{n}}\right)^{n}e^{\sqrt{n}u} \, \mathrm{d}u \right). \tag{1}
\end{align*}
This expression will be our starting point for examining the limiting behavior of $A_n$.
Step 2. To make the analysis simpler, set $f_n, g_n : [0,\infty) \to \mathbb{R}$ by
$$ f_n(u) = \left(1 + \frac{u}{\sqrt{n}}\right)^{n}e^{-\sqrt{n}u}, \qquad g_n(x) = \left(1 - \frac{u}{\sqrt{n}}\right)^{n}e^{\sqrt{n}u}\mathbf{1}_{[0,n]}(u). $$
We will make heavy use of the following observation on $f_n$ and $g_n$:


*

*$f_n(u) \downarrow e^{-u^2/2}$ and $g_n(u) \uparrow e^{-u^2/2}$ as $n\to\infty$ for each fixed $u \geq 0$.


Indeed, this can be easily proved by allowing $n$ to take any positive real value and investigating first two derivatives of $\log f_n$ and $\log g_n$. Now using this,
$$ \sqrt{n} \int_{\sqrt{n}/2}^{\infty} f_n(u) \, \mathrm{d}u
\leq \sqrt{n} \int_{\sqrt{n}/2}^{\infty} f_{1}(u) \, \mathrm{d}u
= \frac{1}{2}\sqrt{n}(\sqrt{n}+4)e^{-\sqrt{n}/2} $$
and
$$ \sqrt{n} \int_{\sqrt{n}/2}^{\sqrt{n}} g_n(u) \, \mathrm{d}u
\leq \sqrt{n} \int_{\sqrt{n}/2}^{\infty} e^{-u^2/2} \, \mathrm{d}u
\leq 2 \int_{\sqrt{n}/2}^{\infty} u e^{-u^2/2} \, \mathrm{d}u
\leq 2 e^{-n/8}.
$$
So we may truncate both integral at $\sqrt{n}/2$ to write
$$ A_n = \sqrt{n} \int_{0}^{\sqrt{n}/2} (f_n(u) - g_n(u)) \, \mathrm{d}u + o(1). $$
Finally, let $0 \leq u \leq \sqrt{n}/2$ and notice that
\begin{align*}
\frac{g_n(u)}{f_n(u)}
&= \exp\left( 2\sqrt{n}u - 2n \sum_{k=1,3,5,\cdots} \frac{u^k}{k n^{k/2}} \right) \\
&\geq \exp\left( - 2n \sum_{k=3,5,7,\cdots} \frac{u^3}{k n^{3/2}} \left(\frac{1}{2}\right)^{k-3} \right) \\
&= \exp\left( - \frac{c u^3}{\sqrt{n}} \right)
\end{align*}
for some absolute constant $c \in (0, \infty)$, and so,
$$ \sqrt{n}\left(f_n(u) - g_n(u) \right) \leq \sqrt{n} \left( 1 - e^{-c u^3/\sqrt{n}} \right) f_n(u)
\leq c u^3 f_n(u) $$
uniformly in $0 \leq u \leq \sqrt{n}/2$. (In the last step, we utilized the inequality $1 - e^{-x} \leq x$.) Therefore by the dominated convergence theorem,
\begin{align*}
\lim_{n\to\infty} A_n
&= \int_{0}^{\infty} \lim_{n\to\infty} \sqrt{n}\left(f_n(u) - g_n(u) \right) \mathbf{1}_{[0,\sqrt{n}/2]}(u) \, \mathrm{d}u \\
&= \int_{0}^{\infty} \frac{2}{3}u^3 e^{-u^2/2} \, \mathrm{d}u \\
&= \frac{4}{3}
\end{align*}
as desired. $\square$
A: Write the inner sums as
$2\sum_{k=0}^{n}\frac{n^k}{k!}-e^n
$
and look up Ramanujan.
(Done on my phone so I can't do much more right now.)
