Calculate limit with summation index in formula [duplicate]

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'.

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