Double Limits!

Changing variables, we can write $L(n) \; = \; \frac{e^n}{n^{n+\frac12}}\int_n^\infty y^n e^{-y}\,dy \; = \; \frac{e^n}{n^{n+\frac12}}\Gamma(n+1,n)$ using the incomplete Gamma function. But $\frac{\Gamma(n+1,n)}{\Gamma(n+1)} \; \sim \; \tfrac12 \qquad \qquad n \to \infty$ and hence (using Stirling's approximation) $L(n) \; \sim \; \sqrt{\frac{\pi}{2}} \qquad \qquad n \to \infty$ Thus $L \,=\, \sqrt{\frac{\pi}{2}}$ . Since $\pi$ is transcendental, so is $L$ , making the answer $\lfloor 1000 L \rfloor \,=\, \boxed{1253}$ .

The following is an informal (and imprecise) but elementary explanation as why the result makes sense. For a fixed $n$ , expand the integrand binomially, integrate term-by-term and use the result (which may be derived by integrating by-parts) that for any non-negative integer $k$ : $\int_{0}^{\infty} \exp(-x)x^k dx = k!$ Hence the integral is equal to $I_n= \sum_{i=0}^{n} \prod_{k=0}^{i} (1-\frac{k}{n})$ Now, we will approximate this sum term-by-term. Note that, for "small" $k/n$ , we have $\exp(-k/n) \approx 1-\frac{k}{n}$ . Thus, we may write $I_n \approx \sum_{i=0}^{n} \exp(-\sum_{k=0}^{i} k/n) \approx \sum_{i=0}^{n} \exp(-i^2/2n) \approx \int_{0}^{n} \exp(-x^2/2n) dx$ Thus, $\frac{I_n}{\sqrt{n}} \approx \frac{1}{\sqrt{n}} \int_{0}^{n} \exp(-x^2/2n) dx = \int_{0}^{\sqrt{n}} \exp(-z^2/2) dz \to \int_{0}^{\infty} \exp(-z^2/2) dz= \sqrt{\frac{\pi}{2}}$