Evaluate the limit:
$$
\lim_{n \to \infty}e^{-n}\sum_{k = 0}^n \frac{n^k}{k!}
$$
It is not as easy as it seems and the answer is definitely not 1.
Please help in solving it.
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Arthur
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Divyansh Garg
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1can you typeset rather than posting an image? – Troy Woo Aug 26 '14 at 17:08
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1A first step would be to guess the limit. What could it be? – Damien L Aug 26 '14 at 17:14
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Perhaps you could use the Stolz Cesaro Lemma – rehband Aug 26 '14 at 17:19
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How do you know "the answer definitely is not 1"? – Timbuc Aug 26 '14 at 17:20
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See this question – JimmyK4542 Aug 26 '14 at 17:23
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Thanks for the link. I was not aware that it was a duplicate. – Divyansh Garg Aug 26 '14 at 17:36
1 Answers
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Given an event whose frequencies open the Poisson distribution and occurs an average of $n$ times per trial, the probability that it occurs $k$ times in a given trial is
$e^{-n} \frac{n^k}{k!}$.
So, the sum in the limit is the probability that the event (which now must have an integer average) occurs no more than the mean number of times. For large $n$, the Poisson distribution is well-approximated by the normal distribution (this can be made into a precise limiting statement). The normal distribution is symmetric about its mean, so the limit of the sum is the probability that a normally distributed random variable is less than the mean of the variable, namely $\frac{1}{2}$.
Travis Willse
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