# A limit involves series and factorials

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

• 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 math.stackexchange.com/questions/160248 will provide a hint. – GEdgar Feb 3 '13 at 2:16

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:)