Evaluating $\sum_{r=1}^n \frac{\tan(x/2^r)}{2^{r-1}\cos(x/2^{r-1})}$

I was asked to find the sum of $$\sum_{r=1}^n \dfrac{\tan \dfrac{x}{2^r}}{2^{r-1}\cos\dfrac{x}{2^{r-1}}}$$

I proceeded by breaking $$\tan x$$ into $$\sin x$$ and $$\cos x$$ and writing numerator as follows $$\sin\left(\frac{x}{2^{r-1}}-\frac{x}{2^r} \right)$$ then by opening brackets and simplifying I got $$\frac{1}{2^{r-1}}\left(\tan\frac{x}{2^{r-1}}-\tan\frac{x}{2^r}\right)$$ But I couldn't proceed from here.

Any help will be appreciated

• Write a few values of $r$ to recognize en.wikipedia.org/wiki/Telescoping_series – lab bhattacharjee Jul 22 at 7:31
• It was able to solve it using telescopic sum if $\frac{1}{2^{r-1}}$was not attached to it.But i am unable to apply it in this question. – utkarsh bhatt Jul 22 at 7:53
• Mathematica gives a result equivalent to $$\sum _{r=1}^n \frac{1}{2^{r-1}}\left( \tan \left(\frac{x}{2^{r-1}}\right)-\tan \left(\frac{x}{2^r}\right)\right) =2 \csc (2 x)-2^{1-n} \csc \left(2^{1-n} x\right),$$ but unfortunately I can't find a way to get there. – James Arathoon Jul 22 at 9:26
• Very SORRY everybody for my goofings, now you may see the fully correct answer of mine. – Dr Zafar Ahmed DSc Jul 22 at 12:24
• @utkarshbhatt, What's the actual term under summation as it differs with the highest voted post – lab bhattacharjee Jul 23 at 11:23

Start with $$\cos(x/2) \cos(x/2^2) \cos(x/2^3)....\cos(x/2^n)= \frac{\sin x}{2^n\sin(x/2^{n})}$$ Take $$\ln$$ of both sides, then $$\sum_{r=1}^{n} \ln \cos (x/2^{r}) = -\ln \sin (x/2^{n})+\ln \sin x-n\ln 2$$ D.w.r.t x and get $$\sum_{r=1}^{n} \frac{1}{2^{r}} \tan (x/2^{r})=\frac{1}{2^{n}} \cot (x/ 2^{n})- \cot x=g(x) ~~~~(1)$$ Similarly $$\cos x \cos(x/2) \cos(x/2^2) \cos(x/2^3)....\cos(x/2^{n-1})= \frac{\sin 2x}{2^n\sin(x/2^{n-1})}$$ Take $$\ln$$ of both sides, then $$\sum_{r=1}^{n} \ln \cos (x/2^{r}) = -\ln \sin (x/2^{n-1})+\ln \sin 2x-n\ln 2$$ D.w.r.t x and get $$\sum_{r=1}^{n} \frac{1}{2^{r-1}} \tan (x/2^{r-1})=\frac{1}{2^{n-1}} \cot (x/ 2^{n-1})- 2\cot 2 x=f(x) ~~~~(2)$$ The required sum is $$S=\sum_{r=1}^{n} \frac{1}{2^{r-1}} [\tan (x/2^{r-1})- \tan (x/2^r)] = f(x)-2g(x)$$ $$\Rightarrow S =\frac{1}{2^{n-1}} \left( \cot (\frac{x}{2^{n-1}})- \cot(\frac{x}{2^n}) \right) -2 (\cot 2 x -\cot x)$$ $$\Rightarrow S=-2^{1-n} \csc(x2^{1-n})+2\csc(2x).$$

• the correct answer is$$\sum _{r=1}^n \frac{1}{2^{r-1}}\left( \tan \left(\frac{x}{2^{r-1}}\right)-\tan \left(\frac{x}{2^r}\right)\right) =2 \csc (2 x)-2^{1-n} \csc \left(2^{1-n} x\right)$$ – utkarsh bhatt Jul 22 at 11:44
• how is g(x) = g(x) -2f(x) – utkarsh bhatt Jul 22 at 11:48
• @James Arathoon Thanks you were right – Dr Zafar Ahmed DSc Jul 22 at 12:26
• @Utkarsh bhatt you were right and $g(x)=f(x)-2g(x)$ was a typo in the first version of my answer. – Dr Zafar Ahmed DSc Jul 22 at 12:27
• Very Sorry for my goofings, you may see the correct answer now. – Dr Zafar Ahmed DSc Jul 22 at 12:28

$$\sum^{n}_{k=1}\frac{\tan(x/2^k)}{2^{k-1} \cos(x/2^{k-1})} = \sum^{n}_{k=1}\frac{\sin^2(x/2^k)}{2^{k-1}\sin (x/2^k)\cos(x/2^k)\cos(x/2^{k-1})}$$

$$=2\sum^{n}_{k=1}\frac{1-\cos(x/2^{k-1})}{2^{k-1}\sin(x/2^{k-1})\cos(x/2^{k-1})}$$ $$= 2\sum^{n}_{k=1}\bigg[\frac{1}{2^{k-2}\sin(x/2^{k-2})}-\frac{1}{2^{k-1}\sin(x/2^{k-1})}\bigg]$$

$$= \bigg[\frac{2}{\sin x}-\frac{2}{2^n \sin (2x/2^n)}\bigg].$$

• I think there is a calculation mistake in the last line of your answer.Anyway thanks for the approach, but is there a way to guess which term you should multiply to get terms cancelled out. – utkarsh bhatt Jul 22 at 17:32
• @Utkarsh Bhatt the expected answer should have $[\frac{2}{\sin 2x}...$ – Dr Zafar Ahmed DSc Jul 23 at 16:02