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I am following a course on stochastic processes. One question is to Calculate $$lim_{n\to \infty} \sum^n_{i=0} e^{-n} \frac{n^i}{i!}$$ I see that it looks like the expected value of $e^{-n} \frac{n^{i+1}}{i!}$ but i don't know how to proceed. I tried to use mathematica but i'm not sure this is correct.

Solve[Integrate[E^(-n) n^(i + 1)/Factorial[i], {n, 0, x}] == 1/2 Integrate[ E^(-n) n^(i + 1)/Factorial[i], {n, 0, Infinity}] , x, Reals]

Thanks in advance

Fibonacci
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  • Yea i saw it after posting this, will close it now – Fibonacci May 14 '18 at 10:46
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    I hope that the duplication mark does not discourage you; it simply means that this community has accumulated enough knowledge to readily answer your question. Especially, the user Did's answer there provides a probabilistic solution. – Sangchul Lee May 14 '18 at 10:52
  • Note that to make the argument formal you might want to consider probability generating functions/moment generating functions/characteristic functions, to show that the sum of $n$ i.i.d. $\mathrm{Pois}(1)$ random variables is indeed $\mathrm{Pois}(n)$ distributed. – Václav Mordvinov May 14 '18 at 10:59

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Hint: Consider a sequence of independent and identically distributed $\mathrm{Pois}(1)$ distributed random variables, and apply the central limit theorem. The value should be $\frac12$.

Václav Mordvinov
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