Question

Value of $\lim_{n \rightarrow \infty} e^{-n}\left(1+n+\frac{n^{2}}{2 !}+\dots+\frac{n^{n}}{n !}\right)$ is

• A. $0$
• B. $1$
• C. $1/2$
• D. $\infty$

Hint:

This is a famous result in probability. The sum of the first $n+1$ terms of the Taylor series for $e^n$, divided by $e^n$, always tends to $0.5$ because $n$ is the mean (and nearly the median) of the $Poisson(n)$ distribution.

Answer 1

The Correct Option is C

This limit relates to the median of a Poisson distribution as the parameter goes to infinity.

Step 1: \color{redFormulate as a Poisson probability
Let $X_n \sim Poisson(\lambda=n)$.
The probability mass function is $P(X_n = k) = \frac{e^{-n} n^k}{k!}$.
The given sum is $P(X_n \le n) = \sum_{k=0}^{n} \frac{e^{-n} n^k}{k!}$.

Step 2: \color{redApply CLT to Poisson
A Poisson variable with parameter $n$ can be viewed as the sum of $n$ independent $Poisson(1)$ variables.
By CLT, as $n \to \infty$, $X_n \approx N(\mu=n, \sigma^2=n)$.

Step 3: \color{redEvaluate the limit
$P(X_n \le n) = P\left(\frac{X_n - n}{\sqrt{n}} \le \frac{n - n}{\sqrt{n}}\right) = P(Z_n \le 0)$.
As $n \to \infty$, $Z_n \to Z \sim N(0,1)$.
$P(Z \le 0) = 1/2$.
Thus, the limit is $1/2$.