Can $\prod\limits_{k=1}^\infty \left(1- \frac{1}{e^{ \sqrt{2} \pi k}}\right)$ be put into closed form? Let $$ \alpha =  \prod_{k=1}^{\infty} \left(1- \frac{1}{e^{ \sqrt{2} \pi  k}}\right) $$
and $$ \beta = \prod_{k=1}^{\infty} \left(1 + \frac{1}{e^{ \sqrt{2} \pi  k}}\right) = \frac{\exp \left(\frac{\sqrt{2}\pi }{24}\right)}{\sqrt[4]{2}} $$
Then $$ \sum_{n=1}^{\infty} \text{arctanh} \left(\frac{1}{e^{ \sqrt{2} \pi  n}} \right) = -\text{arctanh} \left( \frac{\alpha - \beta}{\alpha + \beta}\right) \approx 0.0119025137323901862  $$
Is there a way to solve for the product $\alpha$? I figure that it's likely given that $\beta$ can be.
 A: Let $q=e^{-\pi\sqrt{2}}$ and then the product $\alpha$ in question is $$\alpha=\prod_{n=1}^{\infty}(1-q^n)\tag{1}$$ This is related to the famous Dedekind eta function given by $$\eta(q) =q^{1/24}\prod_{n=1}^{\infty}(1-q^n)\tag{2}$$ which in turn has a closed form in terms of elliptic modulus $k$ associated with nome $q$ and complete elliptic integral of first kind $K(k) $: $$\eta(q) =2^{-1/6}\sqrt{\frac{2K(k)}{\pi}}k^{1/12}k'^{1/3}\tag{3}$$ where
\begin{align}
k&=\frac{\vartheta_{2}^2(q)}{\vartheta_3^2(q)}\tag{4a}\\
k'&=\sqrt{1-k^2}\tag{4b}\\
K(k)&=\int_0^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}}\tag{4c}\\
\vartheta_2(q)&=\sum_{n\in\mathbb{Z}}q^{(n+(1/2))^2}\tag{4d}\\
\vartheta_3(q)&=\sum_{n\in\mathbb{Z}}q^{n^2}\tag{4e}
\end{align}
Both Jacobi and Ramanujan established that if $p$ is a positive rational number and $q=e^{-\pi\sqrt{p}} $ then the modulus $k$ in $(4a)$ is an algebraic number. Chowla and Selberg further proved (in this paper) that for such $q$ and corresponding $k$ the elliptic integral $K(k) $ can be expressed in closed form containing $\pi$ and values of Gamma function at rational points.
The evaluation of $k$ for $p=2,q=e^{-\pi\sqrt{2}}$ is done via modular equation of degree $2$ and one can show that $$k=\sqrt{2}-1,k' =\sqrt{2(\sqrt{2}-1)}$$ The value of $K(k) $ is given by $$K(k) = \frac{\sqrt{\sqrt{2} +1} \Gamma (1/8)\Gamma (3/8)}{2^{13/4}\sqrt{\pi}} $$ and is evaluated in this answer. Using these values the value of $\eta(q) $ in $(2)$ is obtained and then $\alpha =q^{-1/24}\eta(q)$ gets evaluated in closed form.
The product $\beta$ is simpler to handle because we have $$\beta=q^{-1/24}\cdot\frac{\eta(q^2)}{\eta(q)}\tag{5}$$ and we have $$\eta(q^2)=2^{-1/3}\sqrt {\frac{2K(k)}{\pi}}(kk')^{1/6}\tag {6}$$ and thus $\beta$ is $e^{\pi\sqrt {2}/24}$ times an algebraic number (which turns out to be $2^{-1/4}$ using given values of $k, k'$).
