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Find the following limit:
[tex]\lim_{n\rightarrow +\infty} e^{-n} \sum_{k=0}^n \frac{n^k}{k!}[/tex]
[tex]\lim_{n\rightarrow +\infty} e^{-n} \sum_{k=0}^n \frac{n^k}{k!}[/tex]
micromass said:Find the following limit:
[tex]\lim_{n\rightarrow +\infty} e^{-n} \sum_{k=0}^n \frac{n^k}{k!}[/tex]
mfb said:Well, 1 is an upper limit:
$$\lim_{n\rightarrow +\infty} e^{-n} \sum_{k=0}^n \frac{n^k}{k!} \leq \lim_{n\rightarrow +\infty} e^{-n} \sum_{k=0}^\infty \frac{n^k}{k!} = \lim_{n\rightarrow +\infty} 1 = 1$$
It is interesting that
$$\lim_{n\rightarrow +\infty} e^{-n} \sum_{k=0}^n \frac{n^k}{k!} = \lim_{n\rightarrow +\infty} e^{-n} \sum_{k=0}^{n-1} \frac{n^k}{k!}$$ but I am not sure how to convert this to a lower limit on the limit.
Edit: Okay, that approach does not work. And the limit is not 1. I guess 1/2.
The question can be re-phrased as "in a Poisson distribution with expectation value n, what is the probability to get at most the expectation value?" - and then the limit of n->infinity. As the Poisson distribution approaches a Gaussian distribution, I would expect 1/2 as limit.
lurflurf said:You really like Obfuscation.
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