Trigonometric series sum involving tangents 
$$\frac{1}{4}\tan\bigg(\frac{\pi}{8}\bigg)+\frac{1}{8}\tan\bigg(\frac{\pi}{16}\bigg)+\frac{1}{16}\tan\bigg(\frac{\pi}{32}\bigg)+\cdots\cdots \infty$$

Try: $$\cos x\cdot \cos(x/2)\cdot\cos(x/2^2)\cdots\cdots \cos(x/2^{n-1})=\frac{1}{2^{n-1}}\frac{\cos(x)}{\sin(x/2^{n-1})}$$
Could some help me how to solve ahead , thanks
 A: Using the identity:
$$
\cot x-\tan x = 2\cot 2x \Rightarrow \tan x=\cot x -2\cot 2x,
$$
one obtains a telescoping sum:
$$
S(x):=\sum_{n=0}^\infty \frac{\tan(x/2^n)}{2^n}=\sum_{n=0}^\infty\left( \frac{\cot(x/2^n)}{2^n}-\frac{\cot(x/2^{n-1})}{2^{n-1}}\right)\\=
-2\cot(2x)+\lim_{n\rightarrow\infty}\frac{\cot(x/2^n)}{2^n}=\frac{1}{x}-2\cot (2x).
$$
Applying this to your case one obtains 
$$\frac{1}{4}S\left(\frac{\pi}{8}\right)=\frac{2}{\pi}-\frac{1}{2}.$$
A: Using Viete's infinite product identity 
$$\cos\frac{x}{2}\cdot\cos\frac{x}{4}\cdot...=\prod_{k\in\mathbb{N}}\cos\frac{x}{2^k}=\frac{\sin x}{x}$$
we obtain by taking logarithms on both sides and differentiating with respect to $x$ the following 
$$-\frac{1}{2}\tan\frac{x}{2}-\frac{1}{4}\tan\frac{x}{4}-...=-\sum_{k\in\mathbb{N}}\frac{1}{2^k}\tan\frac{x}{2^k}=\frac{\cos x}{\sin x}-\frac{1}{x}$$
Therefore 
$$\sum_{k\geqslant 2}\frac{1}{2^k}\tan\frac{x}{2^k}=\frac{1}{x}-\frac{\cos x}{\sin x}-\frac{1}{2}\tan\frac{x}{2}$$
Set $x=\pi/2$ to get
$$\frac{1}{4}\tan\frac{\pi}{8}+\frac{1}{8}\tan\frac{\pi}{16}+...=\sum_{k\geqslant 2}\frac{1}{2^k}\tan\frac{\pi}{2^{k+1}}=\frac{2}{\pi}-\frac{\cos (\pi/2)}{\sin(\pi/2)}-\frac{1}{2}\tan\frac{\pi}{4}=\frac{2}{\pi}-\frac{1}{2}$$
