1. **Problem Statement:** Find the average power of the discrete-time signal $x[n] = \cos\left(\frac{\pi n}{3}\right)$.\n\n2. **Formula for Average Power:** The average power $P$ of a discrete-time signal $x[n]$ is given by:\n$$P = \lim_{N \to \infty} \frac{1}{2N+1} \sum_{n=-N}^N |x[n]|^2$$\n\n3. **Apply the formula:** For $x[n] = \cos\left(\frac{\pi n}{3}\right)$, we have:\n$$|x[n]|^2 = \cos^2\left(\frac{\pi n}{3}\right)$$\n\n4. **Use trigonometric identity:** Recall that:\n$$\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$$\nSo,\n$$\cos^2\left(\frac{\pi n}{3}\right) = \frac{1 + \cos\left(\frac{2\pi n}{3}\right)}{2}$$\n\n5. **Rewrite the average power expression:**\n$$P = \lim_{N \to \infty} \frac{1}{2N+1} \sum_{n=-N}^N \frac{1 + \cos\left(\frac{2\pi n}{3}\right)}{2} = \frac{1}{2} \lim_{N \to \infty} \frac{1}{2N+1} \sum_{n=-N}^N \left[1 + \cos\left(\frac{2\pi n}{3}\right)\right]$$\n\n6. **Separate the sum:**\n$$P = \frac{1}{2} \left[ \lim_{N \to \infty} \frac{1}{2N+1} \sum_{n=-N}^N 1 + \lim_{N \to \infty} \frac{1}{2N+1} \sum_{n=-N}^N \cos\left(\frac{2\pi n}{3}\right) \right]$$\n\n7. **Evaluate the first limit:**\n$$\lim_{N \to \infty} \frac{1}{2N+1} \sum_{n=-N}^N 1 = 1$$\n\n8. **Evaluate the second limit:** The sequence $\cos\left(\frac{2\pi n}{3}\right)$ is periodic with period 3. Its values over one period are:\n- For $n=0$: $\cos(0) = 1$\n- For $n=1$: $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$\n- For $n=2$: $\cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}$\n\nThe average over one period is:\n$$\frac{1 + (-\frac{1}{2}) + (-\frac{1}{2})}{3} = \frac{1 - 1}{3} = 0$$\nSince the sequence is periodic, the average over all $n$ is also 0.\n\n9. **Combine results:**\n$$P = \frac{1}{2} (1 + 0) = \frac{1}{2}$$\n\n**Final answer:** The average power of $\cos\left(\frac{\pi n}{3}\right)$ is $$\boxed{\frac{1}{2}}$$.