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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.


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

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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