How to solve this product?

Question

I was reading the Wolfram article on infinite products again, and the product $$\prod_{n=1}^{\infty} (1 + \frac{1}{n^k})$$ Is solved on there. I was wondering if anyone knows how to do that? I tried checking the source but couldn't actually find it.

Answer 1

Let $Z = \{ \omega \in \mathbb{C} : \omega^k = -1\}$. Then

$$ 1 + \frac{1}{n^k} = \prod_{\omega \in Z} \left(1 - \frac{\omega}{n} \right). $$

Now plugging this back, the partial product is

$$ \prod_{n=1}^{N} \left( 1 + \frac{1}{n^k} \right) = \prod_{\omega \in Z} \prod_{n=1}^{N} \left(1 - \frac{\omega}{n} \right) = \prod_{\omega \in Z} \frac{\Gamma(N+1-\omega)}{\Gamma(1-\omega)N!}. $$

In view of the Stirling's approximation, we know that

$$ \frac{\Gamma(N+1-\omega)}{N!} \sim \frac{1}{N^{\omega}} \quad \text{as} \quad N \to \infty, $$

and since $\sum_{\omega \in Z} \omega = 0$ for $k \geq 2$, we obtain

$$ \prod_{n=1}^{\infty} \left( 1 + \frac{1}{n^k} \right) = \prod_{\omega \in Z} \frac{1}{\Gamma(1-\omega)}. \tag{*} $$

When $k$ is even, then we know that $\omega \in Z$ if and only if $-\omega \in Z$, and in this case, we can group the factor according to this parity and then utilize the Euler's reflection formula to write

$$ \frac{1}{\Gamma(1-\omega)\Gamma(1+\omega)} = \frac{1}{\omega \Gamma(1-\omega)\Gamma(\omega)} = \frac{\sin(\pi \omega)}{\pi \omega}. $$

You may plug this back to simplify the product $\text{(*)}$.