A limit involves series and factorials Evaluate :
$$\lim_{n\to \infty }\frac{n!}{{{n}^{n}}}\left( \sum\limits_{k=0}^{n}{\frac{{{n}^{k}}}{k!}-\sum\limits_{k=n+1}^{\infty }{\frac{{{n}^{k}}}{k!}}} \right)$$

Ramanujan proved 
(in S. RAMANUJAN, J. Ind. Math. Soc. 3 (1911), 128; ibid. 4 (1911), 151-152; Collected Papers
(Chelsea, New York; 1962), 323-324)
that 
$$e^n/2 = \sum_{k=0}^{n-1} n^k/k! + (n^n/n!) r(n)$$
where, for large $n$,
$r(n) \approx 1/3 + 4/(135n) + O(1/n^2)$.
I found this in
http://journals.cambridge.org/download.php?file=%2FPEM%2FPEM2_24_03%2FS0013091500016503a.pdf&code=fd828d6902ca6a380244640216120c97
via a Google search for 
"ramanujan exponential series" -
I read Ramanujan's collected works many years ago and remembered this result,
but not its details.
This says that
$\begin{align}
\sum_{k=0}^{n} n^k/k!
&\approx e^n/2 + n^n/n!
-(n^n/n!)r(n) \\
&= e^n/2 + (n^n/n!)(1-r(n))
\end{align}
$
Also,
$\begin{align}
\sum_{k=n+1}^{\infty} n^k/k!
&= e^n - \sum_{k=0}^{n} n^k/k!\\
&= e^n - (e^n/2 + (n^n/n!)(1-r(n)))\\
&= e^n/2 - (n^n/n!)(1-r(n))
\end{align}
$
so 
$\begin{align}
\sum_{k=0}^{n} n^k/k! - \sum_{k=n+1}^{\infty} n^k/k!
&\approx (e^n/2 + (n^n/n!)(1-r(n)))
- (e^n/2 - (n^n/n!)(1-r(n)))\\
&= (n^n/n!)(2-2r(n))
\end{align}
$
and
$\begin{align}
(n!/n^n)\left(\sum_{k=0}^{n} n^k/k! - \sum_{k=n+1}^{\infty} n^k/k! \right)
&\approx 2-2r(n) \\
&\to 2-2/3 \\
= 4/3
\end{align}
$.
GEdgar is right! Good guess:)
