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The following exercise was given to me during a course of probability. I guess that this result can be used to check the Lyapunov condition of the Central Limit Theorem. Useful or not, I need to prove that:

$$\lim_{n \rightarrow \infty} \sum_{k=0}^n e^{-n} \frac{n^k}{k!} = \frac{1}{2}.$$

However, can not figure out how to proceed. Can someone give me a suggestion for the first step? It would be very appreciated!

Thanks!

Trig
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  • My first step would be to factor $e^{-n}$ outside the summation (which is a sum with respect to index $k$, so $n$ is constant within that context). – hardmath Jan 16 '16 at 18:11
  • Your summation can be rewritten in terms of a Upper Incomplete gamma function, whose asymptotics for large $n$ should be known https://en.wikipedia.org/wiki/Incomplete_gamma_function#Upper_incomplete_Gamma_function – Pierpaolo Vivo Jan 16 '16 at 18:13
  • More precisely, it is equal to $\gamma(n+1,n)/n!$ [in Mathematica, Gamma[n + 1, n]/n!] – Pierpaolo Vivo Jan 16 '16 at 18:17

1 Answers1

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The summation can be rewritten in terms of an upper incomplete Gamma function (as per comments above). Using the integral representation of the upper incomplete Gamma function we have:

Numerator (using a Laplace'sapproximation of the integral): $$ \gamma(n+1,n)=\int_0^n dt\ t^n e^{-t}=n^{n+1}\int_0^1 dz\ z^n e^{-n z}=n^{n+1}\int_0^1dz\ e^{n[\ln z-z]}\sim n^{n+1}e^{-n}\sqrt{\pi/(2n)} $$ Denominator (using Stirling's approximation): $$ n!\sim \sqrt{2\pi n}(n/e)^n $$ The ratio goes to $1/2$.

Pierpaolo Vivo
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