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How do we prove that

$$4(3\sqrt2-4)=\prod_{n=1}^{\infty}\left({e^{2\pi(2n-1)}-1\over e^{2\pi(2n-1)}+1}\right)^8\tag1$$

Rewrite as, to keep it simple

Let $a=e^{2\pi(2n-1)}$

$$4(3\sqrt2-4)=\prod_{n=1}^{\infty}\left(a-1\over a+1\right)^8\tag2$$

Take the log

$${1\over 8}\ln(12\sqrt2-16)=\sum_{n=1}^{\infty}\ln\left({a-1\over a+1}\right)\tag3$$

We have this series

$$-\ln\left({x-1\over x+1}\right)={2\over x-1}-{2^2\over 2(x-1)^2}+{2^3\over 3(x-1)^3}-\cdots\tag4$$

$$-{1\over 8}\ln(12\sqrt2-16)=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}{(-1)^{m-1}2^m\over m(a-1)^m}\tag5$$

How do I evaluate this double series?

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This one is similar to this answer of your previous question and you should have guessed it. Based on the definition of Ramanujan's class invariants \begin{align}G_{n} &= 2^{-1/4}e^{\pi\sqrt{n}/24}(1 + e^{-\pi\sqrt{n}})(1 + e^{-3\pi\sqrt{n}})(1 + e^{-5\pi\sqrt{n}})\cdots\tag{1}\\ g_{n} &= 2^{-1/4}e^{\pi\sqrt{n}/24}(1 - e^{-\pi\sqrt{n}})(1 - e^{-3\pi\sqrt{n}})(1 - e^{-5\pi\sqrt{n}})\cdots\tag{2}\end{align} the desired product is equal to $(g_{4}/G_{4})^{8}$. From the linked answer we have $$G_{1} = 1, g_{1} = 2^{-1/8}$$ and then $$g_{4}=2^{1/4}g_{1}G_{1} = 2^{1/4}2^{-1/8} = 2^{1/8}$$ Thus $g_{4}^{8} = 2$. Again we have $$g_{4}^{8}G_{4}^{8}(G_{4}^{8} - g_{4}^{8}) = \frac{1}{4}$$ so that $$2x(x - 2) = \frac{1}{4}$$ where $x = G_{4}^{8}$. Thus $$8x^{2} - 16x - 1 = 0$$ or $$x = G_{4}^{8} = \frac{4 + 3\sqrt{2}}{4}$$ and then the desired product is $$\frac{8}{4 + 3\sqrt{2}} = 4(3\sqrt{2} - 4)$$ We have used the following identities for Ramanujan Class Invariants: $$g_{4n} = 2^{1/4}g_{n}G_{n},\, (g_{n}G_{n})^{8}(G_{n}^{8} - g_{n}^{8}) = \frac{1}{4}\tag{3}$$ Note that the first equation $g_{4n} = 2^{1/4}g_{n}G_{n}$ follows directly by multiplying equations $(1)$ and $(2)$ and the second equation in $(3)$ follows by noting that $G_{n} = (2kk')^{-1/12}, g_{n} = (2k/k'^{2})^{-1/12}$ and $k^{2} + k'^{2} = 1$.

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