# Evaluating $\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{2^k}\tan\frac{x}{2^k}$ [duplicate]

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.

Hint:

$$2\cot 2y=\cot y-\tan y$$

$$\implies\tan y=\cot y-2\cot2y$$

Replace $$y$$ with $$\dfrac x{2^r},0\le r\le n$$

to find $$\sum_{r=0}^n\dfrac1{2^r}\cdot\tan\dfrac x{2^r}=\cot x-2\cot2x+1/2(\cot x/2-\cot x)+\cdots+1/2^n(\cot x/2^n-1/2\cot x/2^{n-1})=1/(2^n)\cot (x/2^n)-2\cot2x$$

Finally set $$n\to\infty$$ and replace $$\dfrac x{2^n}$$ with $$h$$ so that $$h\to0$$

and use

$$\lim_{h\to0}\cos h=1=\lim_h\dfrac{\sin h}h$$

• The first line was all I needed but thanks anyway.
– Sam
Jan 4, 2020 at 13:02

Hints:

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

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

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