How to prove that $\sum_{k=1}^\infty \frac{\log(\tanh(2^k))}{2^k}=\log(e^4-1)-4$?

Question

Desmos apparently suggests that $$\sum_{k=1}^\infty \frac{\log(\tanh(2^k))}{2^k}=\log(e^4-1)-4$$with thirteen digits of accuracy (the maximum that Desmos can give). So I wonder how we can prove this.

Turning $\tanh(2^k)$ into $\frac{\sinh(2^k)}{\cosh(2^k)}$ doesn't really make anything easier. Maybe we could use the product formulae of $\sinh$ and $\cosh$, but then we will have to deal with two double summations.

The inspiration for this problem came when I was trying to solve $$\prod_{k=1}^\infty\left(\frac{e^{2^k}-e^{-2^k}}{e^{2^k}+e^{-2^k}}\right)^{\frac{1}{2^k}}$$Which could be found here

Answer 1

Another approach,

You can derive the identity $$\log(\tanh(x)) = 2\log(1-e^{-2x})-\log(1-e^{-4x})$$

Plugging this into the series will cause it to telescope, giving the desired result.

Answer 2

I would prefer to work with the original product. Fix any real number $x>1$.

Fix $N\in\Bbb N$. $$\large\begin{align}\pi_N:&=\prod_{k=1}^N\left(\frac{x^{2^k}-x^{-2^k}}{x^{2^k}+x^{-2^k}}\right)^{2^{-k}}\\&=\frac{(x^2-x^{-2})^{2^{-1}}}{(x^2+x^{-2})^{2^{-1}}}\cdot\prod_{k=2}^N\left((x^2-x^{-2})\frac{\prod_{j=1}^{k-1}(x^{2^j}+x^{2^{-j}})}{x^{2^k}+x^{-2^k}}\right)^{2^{-k}}\\&=(x^2-x^{-2})^{\sum_{j=1}^N2^{-j}}\cdot(x^{2^N}+x^{-2^N})^{-2^{-N}}\cdot\prod_{k=1}^{N-1}(x^{2^k}+x^{-2^k})^{\sum_{j=k+1}^N2^{-j}-2^{-k}}\\&=(x^2-x^{-2})^{\sum_{j=1}^N2^{-j}}\cdot(x^{2^N}+x^{-2^N})^{-2^{-N}}\cdot\prod_{k=1}^{N-1}(x^{2^k}+x^{-2^k})^{-2^{-N}}\\&=(x^2-x^{-2})^{\sum_{j=1}^N2^{-j}}\cdot x^{-\sum_{j=1}^N2^{j-N}}\cdot\prod_{k=1}^N(1+x^{-2^{k+1}})^{-2^{-N}}\end{align}$$From which it is clear: $$\prod_{k=1}^\infty\left(\frac{x^{2^k}-x^{-2^k}}{x^{2^k}+x^{-2^k}}\right)^{2^{-k}}=\lim_{N\to\infty}\pi_N=(x^2-x^{-2})\cdot x^{-2}\cdot1=1-x^{-4}$$In the case of $x=e$, we get the expected result. One can use the dominated convergence theorem on the final product in the last line.

I highly recommend writing out and simplifying the first few partial products so you can see the patterns develop. If I've dropped any indices anywhere, please let me know.