sum of fractional powers

Question

Let $z=e^{i\theta}$ for some real number $\theta$, i.e. $z$ is complex number on the unit circle. Is there a formula for $z+z^{1/2}+z^{1/2^2}+\dots+z^{1/2^n}$? Here $n$ is a positive integer. I only need a formula that holds for almost every $\theta$.

Answer 1

The sum is

$$\sum_{k=0}^n\cos\left(\frac\theta{2^k}\right)+i\sum_{k=0}^n\sin\left(\frac\theta{2^k}\right).$$

Using Taylor to the first order, this yields

$$\sum_{k=0}^n1+i\sum_{k=0}^n\frac\theta{2^k}=n+1+i\theta\frac{1-\frac1{2^{n+1}}}{1-\frac12}.$$

The second order terms add

$$-\frac12\sum_{k=0}^n\frac{\theta^2}{4^k}=-\frac{\theta^2}2\frac{1-\frac1{4^{n+1}}}{1-\frac14}.$$

The third order

$$-i\frac16\sum_{k=0}^n\frac{\theta^3}{8^k}=-\frac{i\theta^3}6\frac{1-\frac1{8^{n+1}}}{1-\frac18}.$$ And so on.

$$n+1+\sum_{k=1}^n\frac{(i\theta)^k}{k!}\frac{1-\frac1{2^{k(n+1)}}}{1-\frac1{2^k}}.$$