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

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 
$$\frac{1}{4}S\left(\frac{\pi}{8}\right)=\frac{2}{\pi}-\frac{1}{2}.$$
