# Find the value of $\prod_{k=1}^{\infty} \big(1+\frac{1}{k^s}\big)$

I am attempting to find values to the family of products given by $$p(s)=\prod_{k=1}^{\infty} \Bigg(1+\frac{1}{k^s}\Bigg)$$ where $$s\in \mathbb{C}$$ and the real part of $$s$$ is greater than $$1$$. In particular the values of $$p(2)$$ and $$p(3)$$ are $$p(2)=\frac{\sinh{\pi}}{\pi}$$ $$p(3)=\frac{\cosh{\Big(\frac{\sqrt{3}}{2}\pi\Big)}}{\pi}$$ by using Wolfram: Alpha.

Firstly, I would like to know if there is a method which allows $$p(s)$$ to be written in a closed form for any $$s\in \mathbb{C}$$. Otherwise, can someone explain how to find the values of $$p(2)$$ and $$p(3)$$ shown above?

• For $s=2$ see math.stackexchange.com/questions/598805/…. The general theory behind these products is Weierstrass factorization theorem. One should also be able to do $s=2^n$ in general IIRC. – Winther Mar 3 at 19:22
• If you combine Weierstrass product formula for $\sin(\pi x)$ with an explicit factorization of $1 + 1/k^{2^n}$ into factors of the form $1 + c/k^2$ you should end up with a closed form for $s = 2^n$ like $$\prod_{k=1}^\infty\left(1 + \frac{1}{k^{2^n}}\right) = \prod_{k=1}^{2^{n-1}} \frac{\sin(\pi z_k)}{\pi z_k}$$ where $z_k = e^{\pi i\cdot \frac{2k-1}{2^n}}$. – Winther Mar 3 at 19:52
• To save others the search, the Dirichlet series has coefficients oeis.org/A045778 – punctured dusk Mar 3 at 20:05

For integers at least:

Following in the lines of this answer, we can write

$$\prod_{k=1}^{N-1} \frac{P(k)}{k^s},$$

where $$P(x)=(x-\alpha_1)\cdots(x-\alpha_s)$$, as

$$\prod_{j=1}^s \frac{\Gamma(N-\alpha_j)}{\Gamma(N)\Gamma(1-\alpha_j)}.$$

Then, using the lemma also described in that answer that

$$\lim_{N\to\infty} \prod_{i=1}^n \frac{\Gamma(N+\alpha_i)}{\Gamma(N+\beta_i)}=1$$

if $$\sum_{i=1}^N \alpha_i=\sum_{i=1}^N \beta_i$$, the terms involving $$N$$ cancel out in the limit and we get

$$\prod_{j=1}^s \Gamma(1-\alpha_j)^{-1}.$$

Using that $$\Gamma(x)=(x-1)\Gamma(x-1)$$, we can write this as

$$(-1)^{s+1}\prod_{j=1}^s \Gamma(-\alpha_j)^{-1}.$$

This doesn't seem to have a nice closed form (or at least not one WolframAlpha can find for $$s=5$$), but we can evaluate it for $$s=2$$ as follows, by noting that

$$\Gamma(i)\Gamma(-i)=i\Gamma(i)\Gamma(1-i)=\frac{i\pi}{\sin(i\pi)}=\frac{\pi}{\sinh \pi},$$

which gives your result. For $$s=3$$, the product becomes the reciprocal of

\begin{align*} \prod_{\omega^3=1}\Gamma(\omega) &=\Gamma(\omega)\Gamma(\overline{\omega})\\ &=|\Gamma(\omega)|^2\\ &=\frac{|\Gamma(1+\omega)|^2}{|\omega|^2}\\ &=\left|\Gamma\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)\right|^2\\ &=\frac{\pi}{\cosh\left(\frac{\pi\sqrt{3}}{2}\right)}. \end{align*}