Would you help me to drive the following limit:
\begin{equation} \lim_{m \to \infty} \frac{\int_{0}^{m}t^{m-1}e^{-t}dt}{(m-1)!}=\frac{\gamma(m,m)}{\Gamma(m)}, \qquad m \in \mathbb{Z^{+}} \end{equation}
Where $\gamma(s,x)$ is the lower incomplete gamma function, and $\Gamma(s)$ is the ordinary gamma function.
p.s. The simulation shows that to limit goes to $\frac{1}{2}$.
I do not know is the solution using the tools of probability theory valid. And the integral $\int_0^m \frac{t^{m-1}e^{-t}}{(m-1)!}\,dt$ is exactly the probability $$ \mathbb P(X_1+\ldots+X_m \leq m) $$ where $X_1,X_2,\ldots$ are independent and identically distributed random variables with exponential distribution with expected value $\mathbb E[X_1]=1$. Indeed, $S_m=X_1+\ldots+X_m$ has Gamma distribution with the probability density function $$ f_{S_m}(t) = \frac{t^{m-1}e^{-t}}{(m-1)!} \cdot \mathbb 1_{t>0}. $$
Next, $$ \mathbb P(X_1+\ldots+X_m \leq m) = \mathbb P\left(\frac{X_1+\ldots+X_m - m\mathbb E[X_1]}{\sqrt{m}} \leq 0\right) \to \Phi(0)=\frac12 $$ as $m\to\infty$ by Central Limit Theorem. Here $\Phi(x)=\int_{-\infty}^x \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx$ is a cumulative distribution function of standard normal distribution.