# Closed form for $\prod_{n=1}^\infty\sqrt{\tanh(2^n)},$

$$\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$$
where $\tanh(z)=\frac{e^z-e^{-z}}{e^z+e^{-z}}$ is the hyperbolic tangent.

#### Solutions Collecting From Web of "Closed form for $\prod_{n=1}^\infty\sqrt{\tanh(2^n)},$"

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

$$P = \exp \log(1 – x^2) = 1 – x^2 = 1 – e^{-4}.$$

Let
$$f(x)=\prod_{n=0}^\infty\left(1-x^{2^n}\right)^{1/2^n}\tag{1}$$
and
$$g(x)=\prod_{n=0}^\infty\left(1+x^{2^n}\right)^{1/2^n}\tag{2}$$
Then
\begin{align} f(x)\,g(x) &=\prod_{n=0}^\infty\left(1-x^{2^{n+1}}\right)^{1/2^n}\\ &=\prod_{n=1}^\infty\left(1-x^{2^n}\right)^{2/2^n}\\ &=\left(\frac{f(x)}{1-x}\right)^2\tag{3} \end{align}
from which we get
$$\frac{f(x)}{g(x)}=(1-x)^2\tag{4}$$
Note that
$$\prod_{n=1}^\infty\left(1-x^{2^n}\right)^{1/2^n}=\frac{f(x)}{1-x}\tag{5}$$
and
$$\prod_{n=1}^\infty\left(1+x^{2^n}\right)^{1/2^n}=\frac{g(x)}{1+x}\tag{6}$$
Therefore, combining $(4)$, $(5)$, and $(6)$, we get
$$\frac{\displaystyle\prod_{n=1}^\infty\left(1-x^{2^n}\right)^{1/2^n}}{\displaystyle\prod_{n=1}^\infty\left(1+x^{2^n}\right)^{1/2^n}}=1-x^2\tag{7}$$
Plug $x=e^{-2}$ into $(7)$ to get
$$\prod_{n=1}^\infty\tanh(2^n)^{1/2^n}=1-e^{-4}\tag{8}$$