Calculate limit with summation index in formula 
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. 
 A: 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'.
A: 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:
$$ 1 = 2L $$
$$ \frac{1}{2} = L $$
