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

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Do you have any thoughts on the problem? If you've tried something and it didn't work, it would be useful to know for anyone trying to solve it themselves. Also, if this problem came up in a certain context (i.e. as an exercise in a textbook following a chapter on BLAH), then it may be helpful to a potential answerer to know what this context is. – Michael Albanese Feb 3 '13 at 1:54
Is it $\sqrt{2\pi}$ ?? – GEdgar Feb 3 '13 at 2:02
@GEdgar I dont know the answer :( – gauss115 Feb 3 '13 at 2:06
How about $4/3$. – GEdgar Feb 3 '13 at 2:08
Maybe the answers/methods in will provide a hint. – GEdgar Feb 3 '13 at 2:16
up vote 4 down vote accepted

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 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} $


$\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} $


$\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} $


$\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:)

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