<s>[INST] Prove that: $ \sum_{i=0}^{\infty} \frac{\tan \frac{\theta}{2^i}}{2^i}= \frac{1}{\theta} - 2 \cot 2 \theta$ My attempt:
Consider the following series:
$$ S = \sum_{i=0}^n \ln( \sec \frac{x}{2^i} )$$
Notice that $ \lim_{n \to \infty} \frac{dS}{dx}$ is the required sum.
Simplfying S,
$$ S = - \ln \left(  \cos x \cdot \cos \frac{x}{2} ... \cos \frac{x}{2^n} \right)$$
or,
$$ S = - \ln \left(  \frac{ \sin(2x)}{2^{n+1} \sin (\frac{x}{2^n})} \right)= \ln(2^{n+1}) + \ln( \sin \frac{x}{2^n}) - \ln( \sin(2x) )$$
Now,
$$ \frac{dS}{dx}  = \frac{1}{2^n} \cot \frac{x}{2^n} - 2 \cot(2x)$$
The problem I'm having is proving that
$$ \lim_{ n \to \infty} \frac{1}{2^n} \cot \frac{x}{2^n} = \frac{1}{x}$$
 [/INST] For a calculus-free solution, prove by induction $\sum_{i=0}^n2^{-i}\tan\tfrac{\theta}{2^i}=2^{-n}\cot(2^{-n}\theta)-2\cot2\theta$ for $n\ge0$. Since $\tan x=\cot x-2\cot2x$, the left-hand side is a telescoping series.
</s>