How do I compute $\prod_{n=1}^\infty \mathrm{e}^{{\mathrm{i}\pi}/{2^n}}$?

Question

I'm struggling with how to compute the following product: $$\prod_{n=1}^\infty \mathrm{e}^{{\mathrm{i}\pi}/{2^n}} $$

Wolfram Alpha tells me it's $-1$, and I can confirm that it converges since

$$\sum_{n=1}^{\infty}\ln(\mathrm{e}^{{\mathrm{i}\pi}/{2^n}}) = \mathrm{i}\pi\sum_{n=1}^{\infty}\frac{1}{2^n} = \mathrm{i}\pi$$

Also, of course

$$\mathrm{e}^{\mathrm{i}\pi/2^n} = \cos(\frac{\pi}{2^n})+\mathrm{i} \sin(\frac{\pi}{2^n})$$

so I can see that the cos part starts at 0 and goes to 1, and vice versa for the sin part. However, it doesn't seem this formulation helps me compute the product.

I have a basic understanding of computing series but haven't done products like this before. I tried to find a method in my calculus book but couldn't. The MathWorld page lists many examples of converging products but also doesn't really describe a method I can use.

Answer 1

You already have the answer! We have

$$\begin{align} \prod_{n=1}^{\infty}e^{i\pi/2^n}&=e^{\log\left(\prod_{n=1}^{\infty}e^{i\pi/2^n}\right)}\\\\ &=e^{\sum_{n=1}^{\infty}\log\left(e^{i\pi/2^n}\right)}\\\\ &=e^{i\pi\sum_{n=1}^{\infty}(1/2^n)}\\\\ &=e^{i\pi}\\\\ &=-1 \end{align}$$

as expected!

Answer 2

Hint:

More simply $e^{ai} e^{bi} = e^{(a+b)i}$. Try adding up the exponents!

Answer 3

Hint: $$\prod_{n=1}^{\infty}e^{a_n}=\lim_{N\to\infty}\prod_{n=1}^{N}e^{a_n}= \lim_{N\to\infty}e^{\sum_{n=1}^{N}a_n} = e^{\lim_{N\to\infty}\sum_{n=1}^{N}a_n}= e^{\sum_{n=1}^{\infty}a_n}$$

provided these limits exist.

Can you determine $$\sum_{n=1}^{\infty}a_n$$ in this case?