The problem asks us to find the range of $\theta$ such that the infinite series $\sum_{n=1}^{\infty} (1 - \cos\theta - \cos2\theta)^n$ converges, given that $0 \le \theta < 2\pi$. The answer should be in the form of intervals.

1. Problem Description

The problem asks us to find the range of $\theta$ such that the infinite series $\sum_{n=1}^{\infty} (1 - \cos\theta - \cos2\theta)^n$ converges, given that $0 \le \theta < 2\pi$. The answer should be in the form of intervals.

2. Solution Steps

The given series is a geometric series with the first term $a = 1 - \cos\theta - \cos2\theta$ and common ratio $r = 1 - \cos\theta - \cos2\theta$.

A geometric series $\sum_{n=1}^{\infty} r^n$ converges if and only if $|r| < 1$.

Therefore, we need to find the values of $\theta$ for which $|1 - \cos\theta - \cos2\theta| < 1$.

$-1 < 1 - \cos\theta - \cos2\theta < 1$

$-2 < - \cos\theta - \cos2\theta < 0$

$0 < \cos\theta + \cos2\theta < 2$

Using the double angle formula, $\cos2\theta = 2\cos^2\theta - 1$, we have:

$0 < \cos\theta + 2\cos^2\theta - 1 < 2$

$0 < 2\cos^2\theta + \cos\theta - 1 < 2$

Let $x = \cos\theta$. Then we have:

$0 < 2x^2 + x - 1 < 2$

First, consider $2x^2 + x - 1 > 0$.

$2x^2 + x - 1 = (2x - 1)(x + 1) > 0$.

This inequality holds when $x > \frac{1}{2}$ or $x < -1$.

Since $x = \cos\theta$, we have $-1 \le x \le 1$. Thus, $x > \frac{1}{2}$ or $x = -1$.

$\cos\theta > \frac{1}{2}$ or $\cos\theta = -1$.

Next, consider $2x^2 + x - 1 < 2$.

$2x^2 + x - 3 < 0$.

$(2x + 3)(x - 1) < 0$.

$-\frac{3}{2} < x < 1$.

Since $-1 \le x \le 1$, we have $-1 \le x < 1$.

$\cos\theta < 1$.

Combining these two conditions, we have $\frac{1}{2} < \cos\theta < 1$ or $\cos\theta = -1$.

Thus, $\cos\theta > \frac{1}{2}$ and $\cos\theta \ne 1$ or $\cos\theta = -1$.

If $\cos\theta > \frac{1}{2}$ and $\cos\theta \ne 1$, then $0 < \theta < \frac{\pi}{3}$ or $\frac{5\pi}{3} < \theta < 2\pi$.

If $\cos\theta = -1$, then $\theta = \pi$.

Therefore, the solution is $0 < \theta < \frac{\pi}{3}$ or $\frac{5\pi}{3} < \theta < 2\pi$ or $\theta = \pi$.

Thus, we have $0 < \theta < \frac{1}{3}(2\pi)$ or $\frac{5}{5}(1)\pi < \theta < \frac{6}{1}(1)\pi$ with $\pi$ missing.

Combining all restrictions, the final intervals are:

$0 < \theta < \frac{\pi}{3}$ or $\frac{5\pi}{3} < \theta < 2\pi$ or $\theta = \pi$.

For the missing parts, we have:

$0 < \theta < \frac{1}{3}(2\pi)$ or $\frac{5}{5}(1)\pi < \theta < \frac{6}{1}(1)\pi$.

Box 1: 0

Box 2: 2

Box 3: 3

Box 4: 5

Box 5: 5

Box 6: 2

3. Final Answer

$0 < \theta < \frac{2}{3}\pi$ or $\frac{5}{5}\pi < \theta < 2\pi$.

1: 0

2: 2

3: 3

4: 5

5: 5

6: 2

Final Answer: $0 < \theta < \frac{2}{3}\pi$ or $\pi < \theta < 2\pi$.

The original answer contains a mistake, $\cos\theta=-1$ is outside of range.

1:0

2:2

3:3

4:1

5:1

6:2

Final Answer: $0 < \theta < \frac{2}{3}\pi$ or $\pi < \theta < 2\pi$.

If $\cos\theta > \frac{1}{2}$, then $0 \le \theta < \frac{\pi}{3}$ or $\frac{5\pi}{3} < \theta \le 2\pi$

If $\cos\theta < 1$, then $\theta \ne 0$

1:0

2:1

3:3

4:5

5:3

6:2

Final Answer: $0 < \theta < \frac{1}{3}\pi$ or $\frac{5}{3}\pi < \theta < 2\pi$