$\Gamma: (0,1) \rightarrow \mathbb{R}$, where $\Gamma(t)=\int_{0}^{\infty}e^{-x}x^{t-1}dx $.
I've already got this result, separating the integral
i)Let $t>0$. Show that:
$$\Gamma(t)=\int_{0}^{\infty}e^{-x}x^{t-1}dx $$ is finite.
$$\Gamma(t)=\int_{0}^{\infty}e^{-x}x^{t-1}dx=\Gamma(t)=\int_{0}^{1}e^{-x}x^{t-1}dx + \int_{1}^{\infty}e^{-x}x^{t-1}dx$$ .
Now, I would like to get the following result:
ii) $$\Gamma(t)=\lim_{n\to\infty} \int_{0}^{n} \big( 1-\frac{x}{n} \big)^n \cdot x^{t-1}dx$$ Using the following results:
$$\lim_{n\to\infty} \big (1-\frac{x}{n} \big)^n = e^{-x}$$ and $$ \big(1-\frac{x}{n} \big)^n \le \big(1-\frac{x}{n+1} \big)^{n+1}$$
Then:
$$ \big(1-\frac{x}{n} \big)^n \le e^{-x}$$ and $$\big(1-\frac{x}{n} \big)^n \rightarrow e^{-x}$$
And I'm not sure if the following steps are correct:
$$ \big(1-\frac{x}{n} \big)^n \cdot x^{t-1} \le e^{-x}\cdot x^{t-1}$$
And then apply the Dominated Convergence Theorem.
And the last result:
iii) if $t>1$ $$\Gamma(t)\cdot \sum_{n=1}^{\infty} \frac{1}{n^t} =\int_{0}^{\infty} \frac{e^{-x}}{1-e^{-x}} \cdot x^{t-1}dx $$
How could he now get the latest identity?
Try writing both sides of the equality separately. But I didn't get equality