# Evaluate $\frac{2}{\sqrt{2}}\cdot \frac{2}{\sqrt{2+\sqrt{2}}}\cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\cdots$

### Problem

Evaluate the infinite product

$$\frac{2}{\sqrt{2}}\cdot \frac{2}{\sqrt{2+\sqrt{2}}}\cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\cdots$$

### Attempt

For convenience，let's rewrite the limit. Denote$$x_1=\sqrt{2},~~~x_{n+1}=\sqrt{2+x_n}(n=1,2,\cdots),$$ then $$\lim_{n \to \infty}\left(\frac{2}{x_1}\cdot \frac{2}{x_2}\cdots \frac{2}{x_n}\right)$$ is what we want.

It's easy to obtain that，$$\{x_n\}$$ is increasing with a greater $$n$$，and convergent to $$2$$. Hence $$x_1Therefore$$\frac{2}{x_1}>\frac{2}{x_2}>\cdots>\frac{2}{x_n}>1.$$ Can we go on from here?

• This is pretty much Viète's formula Dec 30, 2018 at 10:49
• Thus the result is $\dfrac{\pi}{2}$?@Wojowu Dec 30, 2018 at 10:51

Use $$\cos2A=2\cos^2A-1,$$

$$\dfrac2{\sqrt2}=\dfrac2{2\cos\dfrac\pi4}\text{ and }2+\sqrt2=2\left(1+\cos\dfrac\pi4\right)=4\cos^2\dfrac\pi8$$

$$\implies\dfrac2{\sqrt{2+\sqrt2}}=\dfrac2{2\cos\dfrac\pi8}$$

$$\dfrac1{\cos\dfrac\pi4}\dfrac1{\cos\dfrac\pi8}=\dfrac{2\sin\dfrac\pi8}{\cos\dfrac\pi4\sin\dfrac\pi4}=\dfrac{4\sin\dfrac\pi8}{\sin\dfrac\pi2}=?$$

$$\dfrac1{\cos\dfrac\pi4}\dfrac1{\cos\dfrac\pi8}\dfrac1{\cos\dfrac\pi{16}}=2^3\sin\dfrac\pi{2^4}$$

$$\implies\prod_{r=2}^n\cos\dfrac\pi{2^r}=2^{n-1}\sin\dfrac\pi{2^n}$$

$$\implies\lim_{n\to\infty}\prod_{r=2}^n\cos\dfrac\pi{2^r}=\dfrac\pi2\cdot\lim_{n\to\infty}\dfrac{\sin\dfrac\pi{2^n}}{\dfrac\pi{2^n}}=?$$