How to evaluate $\sum_{k=1}^n\ln\left(2\cos\left(\frac{2\pi\cdot3^k}{3^n+1}\right)+1\right)$

Question

By using wolfram alpha, it seems like that

$$\sum_{k=1}^n\ln\left(2\cos\left(\frac{2\pi\cdot3^k}{3^n+1}\right)+1\right)=0 \text{ for all }n\in\mathbb{N}.$$

But I don't know how to prove this identity. Thank you very much.

Answer 1

Taking exponential of both sides, you want

$$ \prod_{k=1}^n \left(2 \cos\left(\dfrac{2\pi\cdot 3^k}{3^n+1}\right)+1\right) = 1 $$

Now note that $1 + 2 \cos(x) = \sin(3x/2)/\sin(x/2)$

and the product telescopes to become

$$\dfrac{\sin \left( \dfrac{3^{n+1}\pi}{3^n+1}\right)}{\sin \left( \dfrac{3 \pi}{3^n+1}\right)} = \dfrac{\sin \left( 3\pi - \dfrac{3\pi}{3^n+1}\right)}{\sin \left( \dfrac{3 \pi}{3^n+1}\right)} = 1$$