A calculus problem by Brian Moehring

Note that $A_n \; =\; e^{-n} \sum_{k=0}^{n + \lfloor \sqrt{n} \rfloor} \frac{n^k}{k!} \; = \; \mathbb{P}[N_n \le n + \sqrt{n}]$ where $N_n$ is a random variable with Poisson distribution $\mathrm{Po}(n)$ . But then we can write $N_n \; = \; X_1 + X_2 + X_3 + \cdots + X_n$ where $X_1,X_2,X_3,\ldots,X_n$ are independent and identically distributed, all with the Poisson distribution $\mathrm{Po}(1)$ . The Central Limit Theorem now tells us that $\tfrac{1}{\sqrt{n}}(N_n - n) \, \to \, Y$ in distribution as $n \to \infty$ , where $Y \sim N(0,1)$ has the standard Normal distribution, and so $\lim_{n \to \infty}A_n \; = \; \lim_{n \to \infty}\mathbb{P}[N_n \le n + \sqrt{n}] \; = \; \lim_{n \to \infty} \mathbb{P}\big[\tfrac{1}{\sqrt{n}}(N_n - n) \le 1\big] \; = \; \mathbb{P}[Y \le 1] \; = \; \Phi(1)$ making the answer $\boxed{0.841344746}$ .