Solving the Limit of the Poisson Sum

Understanding the Sum

The problem asks for the limit: $ L = \lim_{n \to \infty} \sum_{k=0}^{n} \frac{e^{-n} n^k}{k!} $ The summation term $ \frac{e^{-n} n^k}{k!} $ is the probability mass function (PMF) of a Poisson distribution with parameter $\lambda = n$. The sum $ \sum_{k=0}^{n} \frac{e^{-n} n^k}{k!} $ represents the cumulative distribution function (CDF) of a Poisson($n$) random variable, evaluated at $k=n$. Let $X_n \sim Poisson(n)$. The sum is $P(X_n \le n)$.

Applying Normal Approximation

For large values of $n$, the Poisson distribution $Poisson(n)$ can be approximated by a Normal distribution $N(\mu, \sigma^2)$ where the mean $\mu = n$ and the variance $\sigma^2 = n$. This is a consequence of the Central Limit Theorem. So, as $n \to \infty$, $X_n$ behaves like a normally distributed random variable $Y \sim N(n, n)$. We need to find the limit of $P(X_n \le n)$ as $n \to \infty$, which approximates $P(Y \le n)$ where $Y \sim N(n, n)$.

Calculating the Probability

To calculate $P(Y \le n)$ for $Y \sim N(n, n)$, we standardize the variable. Let $Z = \frac{Y - \mu}{\sigma} = \frac{Y - n}{\sqrt{n}}$. $Z$ follows the standard normal distribution $N(0, 1)$. $ P(Y \le n) = P\left(\frac{Y - n}{\sqrt{n}} \le \frac{n - n}{\sqrt{n}}\right) $ $ P(Y \le n) = P\left(Z \le \frac{0}{\sqrt{n}}\right) $ $ P(Y \le n) = P(Z \le 0) $ The probability $P(Z \le 0)$ for a standard normal distribution is the probability of being less than or equal to the mean (which is 0). This value is exactly 0.5.

Conclusion

Therefore, the limit $L$ is 0.5. $ L = \lim_{n \to \infty} P(X_n \le n) \approx P(Z \le 0) = 0.5 $ The value of the limit is 0.5.