<s>[INST] Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$ Please help me to find a closed form for the infinite product
$$\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$$
where $\tanh(z)=\frac{e^z-e^{-z}}{e^z+e^{-z}}$  is the hyperbolic tangent.
 [/INST] For $x < 1$, we have the Taylor series expansion:
$$f(x):= \frac{-1}{4} \log \left(-  \frac{x - x^{-1}}{x + x^{-1}} \right) = \frac{x^2}{2} + \frac{x^6}{6} + \frac{x^{10}}{10} + \frac{x^{14}}{14} + \ldots $$
Then
$$f(x) + \frac{f(x^2)}{2} + \frac{f(x^4)}{4} + \frac{f(x^8)}{8} + \ldots
= \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^6}{6} + \frac{x^8}{8} + \frac{x^{10}}{10} + \ldots  $$
$$= - \frac{1}{2} \log(1 - x^2).$$
Now let $x = e^{-2}$. Then
$$ \log \left( \sqrt[2^n]{\mathrm{tanh}(2^n)} \right) = 
\frac{1}{2^n} \log \left( \frac{e^{2^n} - e^{-2^n}}{e^{2^n} + e^{-2^n}}\right)
$$
$$= \frac{-4}{2^n} f(e^{-2^n}) =  \frac{-4}{2^{n}} f(x^{2^{n-1}}),$$
Hence summing over all $n \ge 1$, we see that, if the product is $P$, then
$$\log P =  -4 \sum_{n=0}^{\infty} \frac{1}{2^{n}} f(x^{2^{n-1}})
= -2 \sum_{n=1}^{\infty} \frac{1}{2^{n}} f(x^{2^{n}}) = \log(1 - x^2),$$
and thus
$$P = \exp \log(1 - x^2) = 1 - x^2 = 1 - e^{-4}.$$
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