The problem is to determine the sum $$\sum_{n=1}^\infty \frac 1{2^n} \tan \frac 1{2^{n+1}}$$
I have been trying everything that comes to my mind (trigonometrical functions' formulas, transforming to integration, Taylor expansion, etc.) to no avail, so a slightest hint to a successful approach would be much appreciated.
I suspect this is a duplicate but I cannot find a question with a complete proof.@ SangchulLee has pointed out he has answered a similar question before. Take a look at that too. In any event, here is my answer. We are going to prove a slightly more general statement:
When $\theta \in (0,\pi)$, all the three numbers $\tan\frac{\theta}{2}$, $\cot\frac{\theta}{2}$ and $\cot\theta$ are defined and satisfy
$$\tan\frac{\theta}{2} = \cot\frac{\theta}{2} - 2\cot\theta$$ Using this, we can convert the sum in $(*1)$ to a telescoping one:
$$\begin{align}\sum_{n=1}^\infty \frac{1}{2^n}\tan\frac{\theta}{2^n} &= \lim_{p\to\infty} \sum_{n=1}^p \frac{1}{2^n}\tan\frac{\theta}{2^n} = \lim_{p\to\infty} \sum_{n=1}^p \left[ \frac{1}{2^n}\cot\frac{\theta}{2^n} - \frac{1}{2^{n-1}}\cot\frac{\theta}{2^{n-1}}\right]\\ &= \lim_{p\to\infty} \left[\frac{1}{2^p}\cot\frac{\theta}{2^p} - \frac{1}{2^{1-1}}\cot\frac{\theta}{2^{1-1}}\right] = \frac{1}\theta \lim_{p\to\infty} \left[\frac{\theta}{2^p}\cot\frac{\theta}{2^p}\right] - \cot\theta \end{align} $$ Since $\lim\limits_{\epsilon\to 0} \epsilon\cot\epsilon = 1$, identity $(*1)$ follows.
For the problem at hand, we can take $\theta = \frac12$ and obtain
$$\sum_{n=1}^\infty \frac{1}{2^n}\tan\frac{1}{2^{n+1}} = 2 - \cot\frac12 \approx 0.169512278287548$$