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.


2 Answers 2



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

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

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

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

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


  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.


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