Evaluating lim\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}

I’m supposed to calculate:

\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}

By using WolframAlpha, I might guess that the limit is \frac{1}{2}, which is a pretty interesting and nice result. I wonder in which ways we may approach it.


The probabilistic way:

This is P[N_n\leqslant n] where N_n is a random variable with Poisson distribution of parameter n. Hence each N_n is distributed like X_1+\cdots+X_n where the random variables (X_k) are independent and identically distributed with Poisson distribution of parameter 1.

By the central limit theorem, Y_n=\frac1{\sqrt{n}}(X_1+\cdots+X_n-n) converges in distribution to a standard normal random variable Z, in particular, P[Y_n\leqslant 0]\to P[Z\leqslant0].

Finally, P[Z\leqslant0]=\frac12 and [N_n\leqslant n]=[Y_n\leqslant 0] hence P[N_n\leqslant n]\to\frac12, QED.

The analytical way, completing your try:

Hence, I know that what I need to do is to find \lim\limits_{n\to\infty}I_n, where

I_n=\frac{e^{-n}}{n!}\int_{0}^n (n-t)^ne^tdt.

To begin with, let u(t)=(1-t)e^t, then I_n=\dfrac{e^{-n}n^n}{n!}nJ_n with

J_n=\int_{0}^1 u(t)^n\mathrm dt.

Now, u(t)\leqslant\mathrm e^{-t^2/2} hence

J_n\leqslant\int_0^1\mathrm e^{-nt^2/2}\mathrm dt\leqslant\int_0^\infty\mathrm e^{-nt^2/2}\mathrm dt=\sqrt{\frac{\pi}{2n}}.

Likewise, the function t\mapsto u(t)\mathrm e^{t^2/2} is decreasing on t\geqslant0 hence u(t)\geqslant c_n\mathrm e^{-t^2/2} on t\leqslant1/n^{1/4}, with c_n=u(1/n^{1/4})\mathrm e^{-1/(2\sqrt{n})}, hence

J_n\geqslant c_n\int_0^{1/n^{1/4}}\mathrm e^{-nt^2/2}\mathrm dt=\frac{c_n}{\sqrt{n}}\int_0^{n^{1/4}}\mathrm e^{-t^2/2}\mathrm dt=\frac{c_n}{\sqrt{n}}\sqrt{\frac{\pi}{2}}(1+o(1)).

Since c_n\to1, all this proves that \sqrt{n}J_n\to\sqrt{\frac\pi2}. Stirling formula shows that the prefactor \frac{e^{-n}n^n}{n!} is equivalent to \frac1{\sqrt{2\pi n}}. Regrouping everything, one sees that I_n\sim\frac1{\sqrt{2\pi n}}n\sqrt{\frac\pi{2n}}=\frac12.

The probabilistic way is shorter, easier, more illuminating, and more fun.

My advice in these matters is, clearly, horribly biased.

Source : Link , Question Author : user 1591719 , Answer Author : Did

Leave a Comment