A limit involves series and factorials

Question

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

Answer 1

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:)