Interesting infinite sum

A simple induction shows that $a_n = 2\cos\big(\tfrac{\pi}{2^{n-1}}\big)$ for all $n \ge 2$ , and so we need to evaluate $I \; = \; \sum_{n=2}^\infty \frac{1}{2^n}\tan\big(\tfrac{\pi}{2^n}\big)$ Note the identity $\cot x - \tan x \equiv 2\cot2x$ . Using this identity, it is a simple induction on $N$ to show that $\sum_{n=2}^N \tfrac{x}{2^n}\tan\big(\tfrac{x}{2^n}\big) \; = \; \tfrac{x}{2^N}\cot\big(\tfrac{x}{2^N}\big) - \tfrac12x\cot\big(\tfrac12x\big) \hspace{2cm} N \ge 2\,,\,0 < x < \pi$ and hence that $F(x) \; = \; \sum_{n=e}^\infty \tfrac{x}{2^n}\tan\big(\tfrac{x}{2^n}\big) \; = \; 1 - \tfrac12x\cot\big(\tfrac12x\big) \hspace{2cm} 0 < x < \pi$ Thus we deduce that $I = \pi^{-1}F(\pi) = \boxed{\pi^{-1}}$ .