Calculating integral using gamma distribution

Question

I've been studying form my Probability theory exam and I found this problem: Calculate using Central limit theorem $$\lim_{n\rightarrow\infty}\int_{0}^{n}\frac{1}{(n-1)!}x^{n-1}e^{-x}dx.$$ Using $$\Gamma(n)=(n-1)!$$ upper expression turns into $$\lim_{n\rightarrow\infty}\int_{0}^{n}\frac{1}{\Gamma(n)}x^{n-1}e^{-x}dx.$$ I know that $$\int_{0}^{n}\frac{1}{\Gamma(n)}x^{n-1}e^{-x}dx=\mathbb{P}(X\leq n )$$ where $$X\sim\Gamma(n,1)$$ How to proceed further from here? I would appreciate any help.

Answer 1

$\frac{1}{\Gamma(n)}x^{n-1} e^{-x}$ is the density of a $\Gamma(n,1)$ distribution, which is the same distribution as the sum of $n$ independent standard exponentials. Thus we have $$ \int_0^n \frac{1}{\Gamma(n)}x^{n-1} e^{-x}dx = P\left(\sum_{i=1}^n X_i \le n\right) = P(\bar X_n\le 1)$$ where $X_i$ are iid standard exponentials and $\bar X_n$ is the sample mean.

The relationship to the CLT should be clear from there.