<s>[INST] Evaluating $\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{2^k}\tan\frac{x}{2^k}$ So I was solving a previous year question paper and stumbled upon the following question-
$${\lim_{n\to {\infty}}}\biggl(\tan x+\frac{1}{2}\tan \frac{x}{2}+\frac{1}{2^2}\tan \frac{x}{2^2}+\frac{1}{2^3}\tan \frac{x}{2^3}+{\cdots}+\frac{1}{2^n}\tan \frac{x}{2^n}\biggl)$$
The only limit with $\tan$ that I have learnt is ${\lim_{x\to 0}}\frac{\tan x}{x}=1$. I have tried the following with no success-
1)Representing $\tan x$ in terms of $\tan \frac x2$   2)Simplifying into $\sin x$ and $\cos x$ 
I do not want a complete solution but would greatly appreciate a hint on how to simplify the series.
 [/INST] Hints:


*

*$\ln\prod\cos\frac{x}{2^n}=\sum\ln\cos\frac{x}{2^n}$

*$\left(\sum\ln\cos\frac{x}{2^n}\right)'=-\sum\frac{1}{2^n}\tan\frac{x}{2^n}$

*$\prod\cos\frac{x}{2^n}$ is something well known.
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