For which $x \in \mathbb C$ does the product $\prod_{n=1}^{\infty} (1+\frac{1}{n})^x(1-\frac{x}{n})$ converge?

Question

For which $x \in \mathbb C$ does the product $\prod_{n=1}^{\infty} (1+\frac{1}{n})^x(1-\frac{x}{n})$ converge?

Concider the product $\prod_{n=1}^{\infty} (1+\frac{1}{n})^x(1-\frac{x}{n})$ where $x \in \mathbb C$. I wanna investigate under which assumptions on $x$ this product converges. I know that a product of the form $\prod_n (1+a_n)$ converges if and only if $\sum_n \log(1+a_n)$ converges but I'm not sure if this can be applied to the product above.

Answer 1

Since $\Gamma(x+1)=\prod_{n\ge1}\frac{(1+1/n)^x}{1+x/n}$, your product is$$\Gamma(x+1)\prod_n(1-x^2/n^2)=\tfrac{1}{\pi}\Gamma(x)\sin(\pi x)=\tfrac{1}{\Gamma(1-x)},$$which is entire.

Answer 2

Using the usual definition of infinite product convergence (meaning the factors are non-zero from some term on and the limit of the product starting from there is non-zero) we can easily show that the product converges normally in the plane (so the limit is an entire function) and is zero precisely where individual factors are zero, hence at positive integers.

Fix a compact set $K$ and pick $n_K >100|x|$ for any $x \in K$ and consider the product starting from $n_K$; since then it is easy to see that the individual terms are less than $3/2$ in absolute values, we can write them as $1+a_n(x), |a_n(x)|<1/2, x \in K, n > n_K$ so using the criterion that $\Pi (1+a_k(x))$ converges normally if $\sum \log(1+a_k(x)$ does, we can estimate and get that

(using that the principal logarithms satisfies $\log (wz)=\log w+\log z$ for $z,w$ close to $1$ since $|(wz-1)-(w-1)-(z-1)| < \pi$ and $\log (wz)$ is about $wz-1$ etc):

$|\log ((1+\frac{1}{n})^x(1-\frac{x}{n}))|=|x\log(1+1/n)+\log(1-x/n)|=O(|x|/n^2+|x|^2/n^2)$

Hence $\sum_{n > n_K} |\log(1+a_k(x)|$ is normally convergent being majorized by $C_K\sum (M/n^2+M^2/n^2), M \ge |x|, x \in K$ hence the product is normally convergent.

As noted the zeroes come only from individual terms only and by inspection we see that they appear at positive integers.