# Find exact value of $\sum_{n=1}^{\infty} \frac{1}{2^{n}} \tan \left( \frac{\pi}{2^{n+1}} \right)$

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?

• Wolfram Alpha doesn't evaluate it exactly, and gets $0.6366197723...$ May 9, 2019 at 1:43
• if it's not $2/ \pi$, then $2/ \pi$ is remarkably close. I wouldn't know how to prove it, though
– Rob
May 9, 2019 at 1:43
• Is there a reason to suspect this has a closed form? May 9, 2019 at 1:45
• 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. May 9, 2019 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.

• 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.
– Yuta
May 9, 2019 at 2:09
• @Yuta Ah yes, that's neat :) May 9, 2019 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 $$x=\pi$$

$$\sum_{n=1}^\infty \frac1{2^n}\tan\left(\fracπ{2^{n+1}}\right) = 2\frac d {dπ}\sum_{n=1}^\infty \ln\left(\sec\left(\fracπ{2^{n+1}}\right)\right) = 2\frac d {dπ}\ln\left(\frac1{\prod_{n=1}^\infty\cos\left(\fracπ{2^{n+1}}\right)}\right)$$ 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 $$2\frac d {dπ}\ln\left(\frac1{\frac2π}\right) = 2\frac d {dπ}\ln\left(\fracπ2\right) = 2\frac{\frac12}{\fracπ2} = \frac2π$$

• (+1) Nice alternative way of calculating the product, haven't seen that before. May 9, 2019 at 2:44
• $\displaystyle \frac{d}{d\pi}???$
– Unit
May 9, 2019 at 2:54
• @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). May 9, 2019 at 3:11
• @Clayton I get it, it's just terrible.
– Unit
May 9, 2019 at 12:20