Question:

Step 1: We are given that $x_1 = \tan^{-1}2 > x_2 > x_3 > \ldots$ are the real numbers satisfying $\sin(x_{n+1} - x_n) + 2^{-(n+1)} \sin x_n \sin x_{n+1} = 0$ for all $n > 1$ and the sequence is convergent. To begin solving this problem, we first use the given condition to derive a relationship between $x_n$ and $x_{n+1}$. Using the trigonometric identity for $\sin(A - B)$, we can rewrite the equation as $\sin x_{n+1} \cos x_n - \cos x_{n+1} \sin x_n + 2^{-(n+1)} \sin x_n \sin x_{n+1} = 0$. Step 2: Rearranging the terms, we get $\sin x_{n+1} (\cos x_n + 2^{-(n+1)} \sin x_n) = \cos x_{n+1} \sin x_n$. Dividing both sides by $\sin x_n \sin x_{n+1}$, we obtain $\frac{\cos x_n + 2^{-(n+1)} \sin x_n}{\sin x_n} = \frac{\cos x_{n+1}}{\sin x_{n+1}}$, which simplifies to $\cot x_n + 2^{-(n+1)} = \cot x_{n+1}$, or $\cot x_{n+1} - \cot x_n = 2^{-(n+1)}$. Step 3: Summing the equation $\cot x_{n+1} - \cot x_n = 2^{-(n+1)}$ telescopically from $n=1$ to $n=k-1$, we have $$ \begin{aligned} \sum_{n=1}^{k-1} (\cot x_{n+1} - \cot x_n) &= \sum_{n=1}^{k-1} 2^{-(n+1)} \\ \cot x_k - \cot x_1 &= \frac{1}{2^2} + \frac{1}{2^3} + \ldots + \frac{1}{2^k} \\ &= \frac{\frac{1}{2^2} (1 - \frac{1}{2^{k-1}})}{1 - \frac{1}{2}} \\ &= \frac{1}{2} (1 - \frac{1}{2^{k-1}}). \end{aligned} $$ Step 4: As $k \to \infty$, the right side converges to $\frac{1}{2}$. Since $x_1 = \tan^{-1}2$, $\cot x_1 = \frac{1}{2}$. Therefore, $\cot x_k$ converges to $\frac{1}{2} + \frac{1}{2} = 1$, implying that $x_k$ converges to $\frac{\pi}{4}$. Step 5: Let $l = \lim_{n \to \infty} x_n$. Then, we have $l = \frac{\pi}{4}$, and $4l = \pi$. Therefore: $\pi$ <div class="key-concept"><strong>Key Concept:</strong> Telescoping series on the cotangent difference reduces the problem to a geometric series that converges to a fixed value.</div> <div class="trap-box"><strong>Trap:</strong> Not recognizing the telescoping nature of $\cot x_{n+1} - \cot x_n = 2^{-(n+1)}$ or miscalculating the geometric series sum.</div>