Show that
\[ \displaystyle \lim_{n \to \infty} \left[ {e}^{n} \sum_{r=0}^{n} \frac{ n^r }{ r! } \right] = \frac{1}{2} \]
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An interesting limit

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Re: An interesting limit
We have also seen this here:
The basic steps are as follows: We consider independent Poisson distributions \(X_1,X_2,\ldots\) with parameter 1. We know that \(Y_n = X_1 + \cdots + X_n\) is Poisson with parameter \(n\). From the central limit theorem, \((Y_nn)/\sqrt{n}\) converges in distribution to the standard normal distribution. In particular, if \(N\) follows a standard normal distribution then \[ \lim_{n \to \infty}\Pr(Y_n \leqslant n) = \lim_{n \to \infty}\Pr\left( \frac{Y_n  n}{\sqrt{n}} \leqslant 0\right) = P(N \leqslant 0) = \frac{1}{2},\] from which we can now read the result.
The basic steps are as follows: We consider independent Poisson distributions \(X_1,X_2,\ldots\) with parameter 1. We know that \(Y_n = X_1 + \cdots + X_n\) is Poisson with parameter \(n\). From the central limit theorem, \((Y_nn)/\sqrt{n}\) converges in distribution to the standard normal distribution. In particular, if \(N\) follows a standard normal distribution then \[ \lim_{n \to \infty}\Pr(Y_n \leqslant n) = \lim_{n \to \infty}\Pr\left( \frac{Y_n  n}{\sqrt{n}} \leqslant 0\right) = P(N \leqslant 0) = \frac{1}{2},\] from which we can now read the result.

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Re: An interesting limit
Thank you for the solution and especially the link, where Mr.Kotronis uses analytic arguments to calculate the given limit! This is exactly what i was looking for... I have heard that the limit can be also calculated using complex methods and i would be happy if someone posts such a solution!
Finally,i would like to add that, in a similar way (using the central limit theorem), one can show that
\[ \displaystyle \lim_{n \to \infty} \left[ \sum_{r=0}^{np} {n \choose r} p^{r} (1p)^{nr} \right] = \frac{1}{2} \]
\[ \displaystyle \lim_{n \to \infty} \left[ \int_{0}^{n} \frac{ {e}^{x}x^{n1} }{\Gamma(n)} \mathrm{d}x \right] = \frac{1}{2} \]
Finally,i would like to add that, in a similar way (using the central limit theorem), one can show that
\[ \displaystyle \lim_{n \to \infty} \left[ \sum_{r=0}^{np} {n \choose r} p^{r} (1p)^{nr} \right] = \frac{1}{2} \]
\[ \displaystyle \lim_{n \to \infty} \left[ \int_{0}^{n} \frac{ {e}^{x}x^{n1} }{\Gamma(n)} \mathrm{d}x \right] = \frac{1}{2} \]