1. ## Nifty series (leads to infinite product of cos)

Hello.

I am asked to calculate the number

$$S = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\sum\limits_{n = 1}^{ + \infty } {\frac{{\tan \left( {\frac{x}{{{2^n}}}} \right)}}{{{2^n}}}} } dx$$

My approach was

$$S = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\sum\limits_{n = 1}^{ + \infty } {\frac{{\tan \left( {\frac{x}{{{2^n}}}} \right)}}{{{2^n}}}} } dx = \sum\limits_{n = 1}^{ + \infty } {\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}} {\frac{{\tan \left( {\frac{x}{{{2^n}}}} \right)}}{{{2^n}}}} dx} = \sum\limits_{n = 1}^{ + \infty } {\ln \left( {\frac{{\cos \frac{\pi }{{3 \cdot {2^{n + 1}}}}}}{{\cos \frac{\pi }{{{2^{n + 1}}}}}}} \right)}$$

$$S = \ln \left( {\frac{{\prod\limits_{n = 1}^{ + \infty } {\cos \frac{\pi }{{3 \cdot {2^{n + 1}}}}} }}{{\prod\limits_{n = 1}^{ + \infty } {\cos \frac{\pi }{{{2^{n + 1}}}}} }}} \right)$$

and I am not actually sure how to further calculate it (computer software shows that equations here are correct and I am getting closer to the right answer).

2. For example we will compute $\prod_{j=1}^{+\infty}\cos \frac{\pi}{2^{j+1}}$.
Let $P_n := \prod_{j=1}^n\cos \frac{\pi}{2^{j+1}}$. What about $\sin \frac{\pi}4P_n$?

3. Originally Posted by girdav
For example we will compute $\prod_{j=1}^{+\infty}\cos \frac{\pi}{2^{j+1}}$.
Let $P_n := \prod_{j=1}^n\cos \frac{\pi}{2^{j+1}}$. What about $\sin \frac{\pi}4P_n$?
Using that logic

$\prod\limits_{j = 1}^{ + \infty } {\cos } \frac{\pi }{{{2^{j + 1}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sin \frac{\pi }{{{2^{n + 1}}}} \cdot \prod\limits_{j = 1}^n {\cos } \frac{\pi }{{{2^{j + 1}}}}}}{{\sin \frac{\pi }{{{2^{n + 1}}}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{{{2^n}}} \cdot \sin \frac{\pi }{2}}}{{\sin \frac{\pi }{{{2^{n + 1}}}}}} = \frac{2}{\pi }$

and I am totally blown away by the elegance of the method

My humble respect, sir.