How does one calculate the following limit using analysis?

Question

I want to evaluate the limit:

$$\lim_{ n \rightarrow +\infty} \frac{1}{\Gamma(n)} \int_0^n x^{n-1}e^{-x}\, {\rm d}x$$

Well $\Gamma$ here stands for the gamma function hence that $\Gamma(n)=(n-1)!$ and of course the integral reminds me of the incomplete Gamma. There is an identity as a limit that says:

$$\lim_{ s \rightarrow +\infty} \frac{\Gamma(s, x)}{\Gamma(s)}=1$$

if I remember that correctly. But I cannot seem to evaluate the limit. I want to see a solution using real analysis methos. I know that it equals $1/2$ but I do not know how to get it.

Answer 1

The usual proof goes along the following lines:

$$ f(x)=\frac{\mathbb{1}_{x>0}}{\Gamma(n)}\cdot x^{n-1}e^{-x} $$ is the probability density function of the sum $X_1+\ldots+X_n$, where the $X_i$s are independent, 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.

A more real-analytic way to prove such claim is to apply Laplace's method (plus Stirling's approximation) to our integral, given that $x=n$ is a stationary point for $g(x)=x^n e^{-x}$.

$$\int_{n}^{+\infty}x^{n}e^{-x}\,dx = n^{n+1}\int_{1}^{+\infty}(xe^{-x})^n\,dx\approx n^{n+1}\int_{1}^{+\infty}\left(\frac{1}{e}e^{-(x-1)^2/2}\right)^n\,dx $$

Answer 2

Using the fact that $~\displaystyle\int_n^\infty\frac{x^n}{e^x}~dx=\frac{n!}{e^n}\cdot\sum_{k=0}^n\frac{n^k}{k!},~$ evaluating the limit is ultimately equivalent to answering this question.