Limited trigonometry

This is just a similar to my problem: Not an usual Trigonometric Summation! .

This is the same solution I had also posted there.

We have (By-multiplying Numerator and Denominator by $\sin\left(\dfrac{x}{2^n}\right)$ ):

$\lim_{n \to \infty} \cos \left(\dfrac{x}{2} \right)\cos \left(\dfrac{x}{2^2} \right)\cos\left(\dfrac{x}{2^3} \right) \ldots \cos \left(\dfrac{x}{2^n} \right) = \dfrac{\sin(x)}{x}$

Differentiating above, we obtain:

$\sum_{n=1}^{\infty} \dfrac{1}{2^n} \tan\left(\dfrac{x}{2^n}\right)=\dfrac1x -\cot(x)$

We'll apply the above result in a step below. Now,

$L = \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{2^k} \tan\left(\dfrac{\pi}{4\cdot2^{k}}\right) = \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{2^k} \tan\left(\dfrac{\pi}{4} \cdot \dfrac{1}{2^k}\right)$

$= \lim_{n \to \infty} \sum_{n=1}^{\infty} \dfrac{1}{2^n} \tan\left(\dfrac{x}{2^n}\right)$ where $x=\dfrac{\pi}{4}$ .

$=\dfrac1x - \cot(x) = \dfrac{4}{\pi} - \cot\left(\dfrac{\pi}{4}\right) =\dfrac{4}{\pi} - 1$

So, $A=4, B=-1, C=-1, A \times B \times C=\boxed{4}$ .

$Let\quad A=\prod _{ r=1 }^{ n }{ cos\left( \frac { x }{ { 2 }^{ r } } \right) } .......(1)\\ First\quad try\quad to\quad prove\quad that\\ A=\frac { 1 }{ { 2 }^{ n } } sinx.cosec\left( \frac { x }{ { 2 } } \right) ...........(2)\\ Taking\quad log\quad on\quad both\quad sides\quad in\quad eqn\quad 1\\ and\quad differentiating\quad wrt\quad x,\\ \frac { { A }' }{ A } ={ { \sum _{ r=1 }^{ n }{ \left( \frac { 1 }{ { 2 }^{ r } } .tan\left( \frac { x }{ { 2 }^{ r } } \right) \right) } } }\\ Now\quad find\quad A'\quad from\quad eqn\quad 2\quad and\quad substitute.\\ Now\quad after\quad substitutions\quad we\quad get\\ \therefore \lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \left( \frac { 1 }{ { 2 }^{ r } } .tan\left( \frac { x }{ { 2 }^{ r } } \right) \right) } } =\frac { 1 }{ x } -cotx\\ \therefore \lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \left( \frac { 1 }{ { 2 }^{ r } } .tan\left( \frac { \frac { \pi }{ 4 } }{ { 2 }^{ r } } \right) \right) } } =\frac { 4 }{ \pi } -1$

JUST USE THE IDENTITY , tan(A) = cot(A) - 2cot(2A) , AND THIS WILL RESULT IN A TELESCOPING SEREIS