Prove that $$ \Gamma\left(z\right) = \lim_{n\to \infty}\int_{0}^{n}t^{z - 1}\left(1 - {t \over n}\right)^{n}\,{\rm d}t \quad\mbox{for}\quad \Re z \gt 0 $$
I know that $$ {\rm e}^{-t/n} = 1 - {t \over n} + {t^{2} \over 2!\,n^{2}} - {t^{3} \over 3!\,n^{3}} + \cdots $$
Then, $$ 0 \leq {\rm e}^{-t/n} - \left(1 - {t \over n}\right) \leq {t^{2} \over 2n^{2}} $$
After there, I cannot do this. Please continue to prove. Thank you for helping.
On
Let be $$\Gamma(z)=\int_0^\infty t^{z-1}\operatorname{e}^{-t}\operatorname{d}t\quad\text{for}\; \Re(z)>0.\tag 1$$ Prove that $$ \Gamma(z)= \lim_{n\to \infty}\int_{0}^{n}t^{z - 1}\left(1 - {t \over n}\right)^{n}\operatorname{d}t \quad\mbox{for}\;\Re(z) > 0.\tag 2 $$
We have to prove that $$ \lim_{n\to \infty}\left[\int_0^\infty t^{z-1}\operatorname{e}^{-t}\operatorname{d}t-\int_{0}^{n}t^{z-1}\left(1 - {t \over n}\right)^{n}\operatorname{d}t\right]=\lim_{n\to \infty}\left[\int_{0}^{n}\left\{\operatorname{e}^{-t}-\left(1 - {t \over n}\right)^{n}\right\}t^{z-1}\operatorname{d}t+\int_n^\infty \operatorname{e}^{-t}t^{z-1}\operatorname{d}t\right] $$ converges to $0$.
From the convergence of the integral on the right in (1) it follows that $$ \lim_{n\to \infty}\int_n^\infty t^{z-1}\operatorname{e}^{-t}\operatorname{d}t=0 $$
Hence we have to prove that $$ \lim_{n\to \infty}\int_{0}^{n}\left\{\operatorname{e}^{-t}-\left(1 - {t \over n}\right)^{n}\right\}t^{z-1}\operatorname{d}t=0 $$
Observe that, when $0\le y< 1$, comparing the series $\operatorname{e}^{y}=1+y+\sum_{k=2}^\infty \frac{y^k}{k!}$ and $\frac{1}{1-y}=1+y+\sum_{k=2}^\infty y^k$ it follows $$ 1+y\le \operatorname{e}^{y}\le \frac{1}{1-y}. $$ Writing $t/n$ for $y$ one has $$ \left(1+\frac{t}{n}\right)^{-n}\ge \operatorname{e}^{-t} \ge \left(1-\frac{t}{n}\right)^{n} $$ and so $$ 0\le \operatorname{e}^{-t} - \left(1-\frac{t}{n}\right)^{n} =\operatorname{e}^{-t}\left[1 - \operatorname{e}^{t}\left(1-\frac{t}{n}\right)^{n}\right] \le \operatorname{e}^{-t}\left[1 - \left(1-\frac{t^2}{n^2}\right)^{n}\right]\tag 3 $$ where we used $\left(1+\frac{t}{n}\right)^{n}\le \operatorname{e}^{t}$ in the last inequality.
Now, if $0\le\alpha\le 1$, $(1-\alpha)^n\ge 1-n\alpha$ by induction when $n\alpha<1$ and obviously when $n\alpha\ge 1$; and, writing $t^2/n^2$ for $\alpha$, we get $$ 1 - \left(1-\frac{t^2}{n^2}\right)^{n}\le \frac{t^2}{n^2}\tag 4 $$ and so from (3) and (4) $$ 0\le \operatorname{e}^{-t}-\left(1 - {t \over n}\right)^{n}\le \operatorname{e}^{-t}\frac{t^2}{n}\tag 5 $$ By the inequalities (5) and by te fact that $|t^z|=t^x$ (where $x$ is the real part of $z$), it follows $$ \left|\int_{0}^{n}\left\{\operatorname{e}^{-t}-\left(1 - {t \over n}\right)^{n}\right\}t^{z-1}\operatorname{d}t\right|\le \int_{0}^{n}\operatorname{e}^{-t}\frac{t^2}{n}t^{x-1}\operatorname{d}t\le \frac{1}{n}\int_{0}^{n}\operatorname{e}^{-t}t^{x+1}\operatorname{d}t. $$ Now $\int_{0}^{\infty}\operatorname{e}^{-t}t^{x+1}\operatorname{d}t$ converges, so $\int_{0}^{n}\operatorname{e}^{-t}t^{x+1}\operatorname{d}t$ is bounded.
Therefore we may conclude that $$ \lim_{n\to \infty}\int_{0}^{n}\left\{\operatorname{e}^{-t}-\left(1 - {t \over n}\right)^{n}\right\}t^{z-1}\operatorname{d}t=0 $$
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\lim_{n \to \infty}\int_{0}^{n}t^{z - 1}\pars{1 - {t \over n}}^{n}\,\dd t = \lim_{n \to \infty}\bracks{n^{z}\int_{0}^{1}t^{z - 1}\pars{1 - t}^{n}\,\dd t} =\lim_{n \to \infty}\bracks{n^{z}\,{\rm B}\pars{z,n + 1}} \\[3mm]&=\lim_{n \to \infty} \bracks{n^{z}\,{\Gamma\pars{z}\,\Gamma\pars{n + 1} \over \Gamma\pars{z + n + 1}}} =\Gamma\pars{z}\,\lim_{n \to \infty}\bracks{n^{z}\, {\Gamma\pars{n + 1} \over \Gamma\pars{z + n + 1}}} \\[3mm]&=\Gamma\pars{z}\,\lim_{n \to \infty}\bracks{n^{z}\, {\pars{n + 1}^{n + 1/2}\expo{-\pars{n + 1}}\root{2\pi} \over \pars{z + n + 1}^{z + n + 1/2}\expo{-\pars{z + n + 1}}\root{2\pi}}} \\[3mm]&=\Gamma\pars{z}\,\lim_{n \to \infty}\bracks{% \pars{n \over n + z + 1}^{z}\pars{n + 1\over n + z + 1}^{n + 1/2}\expo{z}} =\Gamma\pars{z}\expo{z}\lim_{n \to \infty} \pars{1 + {z \over n + 1}}^{-n - 1/2} \\[3mm]&=\Gamma\pars{z}\expo{z}\, \underbrace{\lim_{n \to \infty} \exp\pars{-\bracks{n + \half}\ln\pars{1 + {z \over n + 1}}}} _{\ds{\expo{-z}}} \end{align} where ${\rm B}$ is the $\it Beta$ function. We used the identity $\ds{{\rm B}\pars{x,y} = \int_{0}^{1}t^{x - 1}\pars{1 - t}^{y - 1}\,\dd t}$ and the property $\ds{{\rm B}\pars{x,y} = {\Gamma\pars{x}\Gamma\pars{y} \over \Gamma\pars{x + y}}}$. The final limit is performed by using the Stirling's asymptotic behavior of the Gamma function.