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Possible Duplicate:
Compute the limit: $\lim_{n\rightarrow\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$

I want to calculate the following:

$ \lim_{n \rightarrow \infty} \left( e^{-n} \sum_{i = 0}^{n} \frac{n^i}{i!} \right) $

Numerical calculations show it has a value close to 0.5. But I am not able to derive this analytically. My problem is that I am lacking a methodology of handling the $n$ both as a summation limit and a variable in the equation.

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    According to Wolfram alpha this limit is 1 http://www.wolframalpha.com/input/?2012-12-31

2 Answers 2

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I don't want to put this down as my own solution, since I have already seen it solved on MSE.

One way is to use the sum of Poisson RVs with parameter 1, so that $S_n=\sum_{k=1}^{n}X_k, \ S_n \sim Poisson(n)$ and then apply Central Limit Theorem to obtain $\Phi(0)=\frac{1}{2}$.

The other solution is purely analytic and is detailed in the paper by Laszlo and Voros(1999) called 'On the Limit of a Sequence'.

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    You are welcome2012-12-31
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Well, we can just get rid of $e^{-n}$ rather easily, but that's not what we should do.

$ \lim_{n\rightarrow\infty} e^{-n} \sum_{i=0}^n \frac{n^i}{i!} $

There's something called the Incomplete Gamma Function. It satisfies:

$ \frac{\Gamma(n+1, n)}{n! e^{-n}} = \sum_{k=0}^n \frac{n^k}{k!}$

Substitute:

$ \lim_{n\rightarrow\infty} e^{-n} \frac{\Gamma(n+1, n)}{n! e^{-n}} $

Get rid of $e^{-n}$: $ \lim_{n\rightarrow\infty} \frac{\Gamma(n+1, n)}{n!} $

Now what? Well make a substitution: $ \lim_{n\rightarrow\infty} \frac{\Gamma(n+1, n)}{\Gamma(n+1)} = \frac{1}{2}$

(Note that the following proof might be incorrect, although my CAS agrees with the result and I think it is.)

In order to show this, there is an identity that $\Gamma(a, x) + \gamma(a, x) = \Gamma(a) $, so $\Gamma(a, x) = \Gamma(a) - \gamma(a, x)$. Now find: $ \lim_{n\rightarrow\infty} \frac{\Gamma(n+1) - \Gamma(n+1,x)}{\Gamma(n+1)} $

$ 1 - \lim_{n\rightarrow\infty} \frac{\Gamma(n+1,x)}{\Gamma(n+1)} $

But this is the same as our other limit. If we have: $ \lim_{n\rightarrow\infty} \frac{\Gamma(n+1,x)}{\Gamma(n+1)} = L $

Then: $ 1 - L = L $

So: $ 1 = 2L $ $ \frac{1}{2} = L $

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    @WimC I can try, although I'm not entirely sure if my reasoning is entirely correct. My CAS seems to agree though.2012-12-31