I have to discuss the convergence of the product
$ \prod_{n=1}^{\infty}\Bigg\{ \left(1+\frac{1}{n} \right)^x\left(1-\frac{x}{n}\right)\Bigg\}$
Here is my solution:
Based on the binomial formula, we have
$ \left(1+\frac{1}{n} \right)^x\left(1-\frac{x}{n}\right)=1+\frac{x}{n}+\frac{x(x-1)}{2n^2}+\frac{x(x-1)(x-2)}{6n^3}+\dots -\frac{x}{n}-\frac{x^2}{n^2}-\frac{x^2(x-1)}{2n^3}-\frac{x^2(x-1)(x-2)}{6n^4}-\dots$
$ = 1-\frac{x^2}{n^2}-\frac{x(x-1)}{2n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)$
So we have
$e^{ln C}=\exp\left[\sum_{n=0}^{\infty}\left(\log \left(1-\frac{3x^2-x}{2n^2}+\mathcal{O}\left(\frac{1}{n^3}\right) \right)\right)\right] $
$ \Big|e^{ln C}\Big|\leq\Bigg| \exp\left[\sum_{n=0}^{\infty}\left( -\frac{3x^2-x}{2n^2}+\mathcal{O}\left(\frac{1}{n^3} \right)\right)\right] \Bigg|$
The integral test shows that this series converges, since
$ \int \limits_1^\infty \frac{1}{n^2}\, \mathrm{d}x = \lim_{b \to \infty} \left[-\frac{1}{n}\right]_1^b=1$
Is this right?
Best Answer
Yes, this is mostly right, except that it should be $\int \limits_1^\infty \frac{1}{n^2}\, \mathrm{d}n$. And I'd use some simpler test for that, $1/n^2\le1/n(n-1)$, or condensation test, but that's a matter of taste, naturally. Looking at https://en.wikipedia.org/wiki/Gamma_function#Euler's_definition_as_an_infinite_product , we see that the infinite product is $$1/(-x)!=1/\Gamma(1-x).$$