Evaluate the limit:
$$\lim_{ n \rightarrow +\infty} \frac{1}{\Gamma(n)} \int_0^n x^{n-1}e^{-x}\, {\rm d}x$$
Solution
We have that:
$$ f(x)=\frac{\mathbb{1}_{x>0}}{\Gamma(n)}\cdot x^{n-1}e^{-x} $$
is the probability density function of the sum $X_1+\ldots+X_n$, where the $X_i$s are independent an exponentially distributed ($\lambda=1$) random variables. The original integral is so:
$$ \mathbb{P}[X_1+\ldots+X_n \leq n],\quad\text{where } n = \mathbb{E}[X_1+\ldots+X_n] $$
and the limit equals $\frac{1}{2}$ by the Central Limit Theorem and the result follows.
$$\lim_{ n \rightarrow +\infty} \frac{1}{\Gamma(n)} \int_0^n x^{n-1}e^{-x}\, {\rm d}x$$
Solution
We have that:
$$ f(x)=\frac{\mathbb{1}_{x>0}}{\Gamma(n)}\cdot x^{n-1}e^{-x} $$
is the probability density function of the sum $X_1+\ldots+X_n$, where the $X_i$s are independent an exponentially distributed ($\lambda=1$) random variables. The original integral is so:
$$ \mathbb{P}[X_1+\ldots+X_n \leq n],\quad\text{where } n = \mathbb{E}[X_1+\ldots+X_n] $$
and the limit equals $\frac{1}{2}$ by the Central Limit Theorem and the result follows.
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