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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.

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    $\begingroup$ Hint: $\Gamma(a,1)+\Gamma(b,1)\sim \Gamma(a+b,1)$. So you can write $X=\Gamma(n,1)$ as a sum of... $\endgroup$ – Mike Earnest Jun 23 '18 at 21:20
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    $\begingroup$ So I need to rewrite $\mathbb{P}(X\leq n)$ as $\mathbb{P}(Y_{1}+...+Y_{n}\leq n)$, where $Y_{i}$ are iid random variables such that $Y_{i}\sim\Gamma(1,1)$? $\endgroup$ – Martin Jun 23 '18 at 21:24
  • $\begingroup$ You can use \lim to fix the formatting on $\lim$. $\endgroup$ – joriki Jun 23 '18 at 22:15
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$\frac{1}{\Gamma(n)}x^{n-1} e^{-x}$ 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.

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