Question

Let
\[ L = \lim_{n \to \infty} \sum_{k=0}^{n} \frac{e^{-n} n^{k}}{k!} \]
Which of the following is the value of \(L\)?

• A. 0.5
• B. 1.0
• C. 0
• D. \(e^{-1}\)

Hint:

The sum is the Poisson(n) CDF evaluated at its own mean n; as n grows, the Poisson distribution becomes approximately Normal and symmetric about its mean, so roughly half the probability lies at or below it.

Answer 1

The Correct Option is A

Step 1: Recognize the sum as a Poisson probability.
The term \(\dfrac{e^{-n}n^{k}}{k!}\) is exactly the probability mass function of a Poisson random variable \(N\) with mean \(n\), evaluated at \(k\).
So the full sum \(\sum_{k=0}^{n} \dfrac{e^{-n}n^{k}}{k!}\) is \(P(N \le n)\), the probability that a Poisson(\(n\)) random variable is at most its own mean, \(n\).

Step 2: Use the fact that a Poisson mean and variance are equal.
For \(N \sim Poisson(n)\), both the mean and the variance equal \(n\), so the standard deviation is \(\sqrt{n}\).

Step 3: Apply the Central Limit Theorem as \(n \to \infty\).
A Poisson(\(n\)) variable can be built as the sum of \(n\) independent Poisson(1) variables, so as \(n \to \infty\), the standardized variable \(\dfrac{N-n}{\sqrt{n}}\) converges in distribution to a standard Normal variable, by the Central Limit Theorem.

Step 4: Evaluate the limit at the mean.
\[ P(N \le n) = P\!\left(\frac{N-n}{\sqrt{n}} \le 0\right) \;\to\; \Phi(0) \] as \(n \to \infty\), where \(\Phi\) is the standard Normal CDF.
Since the standard Normal distribution is symmetric about 0, \(\Phi(0) = 0.5\).

Step 5: Rule out the other options.
\(L=1.0\) would only hold if the sum ran all the way to infinity instead of stopping at \(k=n\); stopping exactly at the mean captures only about half the probability mass. \(L=0\) is wrong because a nontrivial chunk (about half) of the Poisson(\(n\)) mass sits at or below its own mean for every \(n\). \(e^{-1}\) is just the single \(k=0\) term of the sum, not the full sum up to \(k=n\).

Final Answer:
\[ \boxed{L = 0.5} \]