TITLE: Find exact value of $\sum_{n=1}^{\infty} \frac{1}{2^{n}} \tan \left( \frac{\pi}{2^{n+1}} \right)$
QUESTION [14 upvotes]: How can I compute the series 
$$\sum_{n=1}^{\infty} \frac{1}{2^{n}} \tan \left( \frac{\pi}{2^{n+1}} \right)$$
I just guess using half-angle formula to compute this series, 
but I can't do any approaches.
How should I do to solve this infinite sum?

REPLY [7 votes]: $$\cot A-\tan A=2\cot2A\implies\dfrac12\tan A=\dfrac12\cot A-\cot2A$$
Using Telescoping series
$$S(m)=\sum_{n=1}^m\dfrac1{2^n}\tan\dfrac x{2^{n+1}}=\dfrac1{2^m}\cot\dfrac x{2^{m+1}}-\dfrac1{2^{1-1}}\cot\dfrac x{2^1}$$
$$\lim_{m\to\infty}S(m)=\dfrac2x-\cot\dfrac x2$$ setting $\dfrac x{2^m}=y,m\to\infty,y\to0$ and $\lim_{y\to0}\dfrac{\sin y}y=1$
Here $x=\pi$