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This question already has an answer here:

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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marked as duplicate by Semiclassical, JimmyK4542, user147263, Antonio Vargas, drhab Aug 26 '14 at 18:20

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    $\begingroup$ can you typeset rather than posting an image? $\endgroup$ – Troy Woo Aug 26 '14 at 17:08
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    $\begingroup$ A first step would be to guess the limit. What could it be? $\endgroup$ – Damien L Aug 26 '14 at 17:14
  • $\begingroup$ Perhaps you could use the Stolz Cesaro Lemma $\endgroup$ – rehband Aug 26 '14 at 17:19
  • $\begingroup$ How do you know "the answer definitely is not 1"? $\endgroup$ – Timbuc Aug 26 '14 at 17:20
  • $\begingroup$ See this question $\endgroup$ – JimmyK4542 Aug 26 '14 at 17:23
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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 $\frac{1}{2}$.

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