How can I compute the series $$\sum_{n=1}^{\infty} \frac{1}{2^{n}} \tan \left( \frac{\pi}{2^{n+1}} \right)$$

I just guess using half-angle formula to compute this series,

but I can't do any approaches.

How should I do to solve this infinite sum?

  • $\begingroup$ Wolfram Alpha doesn't evaluate it exactly, and gets $0.6366197723...$ $\endgroup$ – John Doe May 9 '19 at 1:43
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    $\begingroup$ if it's not $2/ \pi$, then $2/ \pi$ is remarkably close. I wouldn't know how to prove it, though $\endgroup$ – Rob Bland May 9 '19 at 1:43
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    $\begingroup$ Is there a reason to suspect this has a closed form? $\endgroup$ – Clayton May 9 '19 at 1:45
  • $\begingroup$ I would doubt any closed form. There doesn't seem to be any nice way to make it telescope, which would be the only way I would expect it may converge. Numerically computing this isn't so bad though, since the $n$th partial sum has $\mathcal O(4^{-n})$ error. $\endgroup$ – Simply Beautiful Art May 9 '19 at 1:55

Let $$\begin{align}S(x)&=\sum_{n=1}^\infty\frac1{2^n}\tan\left(\frac{x}{2^n}\right)\\&=\frac{d}{dx}\sum_{n=1}^\infty\ln\left(\sec\left(\frac{x}{2^n}\right)\right)\\&=\frac{d}{dx}\ln\left(\prod_{n=1}^\infty\sec\left(\frac{x}{2^n}\right)\right)\\&=\frac{d}{dx}\ln\left(\lim_{m\to\infty}\left\{2^m\csc(x)\sin(2^{-m}x)\right\}\right)\\&=\frac{d}{dx}\ln(x\csc(x))\\&=\frac{(1-x\cot(x))\csc(x)}{x\csc(x)}\\&=\frac1x-\cot(x)\end{align}$$

Then we want to compute $$S(\pi/2)=\frac1{π/2}-0=\bbox[5px,border:2px solid black]{\frac2{\pi}}$$

This agrees with the numerical result on Wolfram Alpha.

Some other details:

I used the step $$\prod_{n=1}^m\sec(2^{-n}x)=2^m\csc(x)\sin(2^{-m}x)$$ in the computation. As Yuta pointed out in the comments, this can be proved by multliplying through by $\csc(2^{-m} x)$ and using $$\begin{align}\csc(2^{-m} x)\prod_{n=1}^m\sec(2^{-n}x)&=\frac1{\sin(2^{-m} x)}\prod_{n=1}^m\frac1{\cos(2^{-n}x)}\\&=2\frac1{\sin(2^{-m+1}x)}\prod_{n=1}^{m-1}\frac1{\cos(2^{-n}x)}\\&=\cdots\\&=2^m\frac1{\sin(x)}\end{align}$$

The limit was evaluated by doing a Taylor expansion.

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    $\begingroup$ For $\prod_{n=1}^m\sec(2^{-n}x)=2^m\csc(x)\sin(2^{-m}x)$, multiplying both sides by $\csc(2^{-m}x)$ and apply the formula $\sin 2\theta=2\sin\theta\cos\theta$ repeatedly should work. $\endgroup$ – Yuta May 9 '19 at 2:09
  • $\begingroup$ @Yuta Ah yes, that's neat :) $\endgroup$ – John Doe May 9 '19 at 2:11

$$\cot A-\tan A=2\cot2A\implies\dfrac12\tan A=\dfrac12\cot A-\cot2A$$

Using Telescoping series

$$S(m)=\sum_{n=1}^m\dfrac1{2^n}\tan\dfrac x{2^{n+1}}=\dfrac1{2^m}\cot\dfrac x{2^{m+1}}-\dfrac1{2^{1-1}}\cot\dfrac x{2^1}$$

$$\lim_{m\to\infty}S(m)=\dfrac2x-\cot\dfrac x2$$ setting $\dfrac x{2^m}=y,m\to\infty,y\to0$ and $\lim_{y\to0}\dfrac{\sin y}y=1$

Here $x=\pi$

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$$\sum_{n=1}^\infty \frac1{2^n}\tan\left(\fracπ{2^{n+1}}\right) = 2\frac d {dπ}\sum_{n=1}^\infty \ln\left(\sec\left(\fracπ{2^{n+1}}\right)\right) = 2\frac d {dπ}\ln\left(\frac1{\prod_{n=1}^\infty\cos\left(\fracπ{2^{n+1}}\right)}\right)$$ But $$\prod_{n=1}^\infty\cos\left(\fracπ{2^{n+1}}\right) = \cos\left(\fracπ4\right)\cos\left(\fracπ8\right)\cos\left(\fracπ{16}\right)\cos\left(\fracπ{32}\right)... = \frac{\sqrt2}2\frac{\sqrt{2+\sqrt2}}2\frac{\sqrt{2+\sqrt{2+\sqrt2}}}2... = \frac2π$$ (This is Viète's formula.) Then our original sum is $$2\frac d {dπ}\ln\left(\frac1{\frac2π}\right) = 2\frac d {dπ}\ln\left(\fracπ2\right) = 2\frac{\frac12}{\fracπ2} = \frac2π$$

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    $\begingroup$ (+1) Nice alternative way of calculating the product, haven't seen that before. $\endgroup$ – John Doe May 9 '19 at 2:44
  • $\begingroup$ $\displaystyle \frac{d}{d\pi}???$ $\endgroup$ – Unit May 9 '19 at 2:54
  • $\begingroup$ @Unit: think of $\pi$ as a variable and differentiate with respect to it (this part is essentially the same as John Doe's answer...the evaluation of the product is where the two answers really differ). $\endgroup$ – Clayton May 9 '19 at 3:11
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    $\begingroup$ @Clayton I get it, it's just terrible. $\endgroup$ – Unit May 9 '19 at 12:20

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