Step-by-step Solution:

\[ \textbf{Let us evaluate the infinite sum cleanly and show why it telescopes.} \] \[ \textbf{Given expression} \] \[ \sin^{-1}\!\left(\frac{1}{\sqrt{2}}\right) + \sin^{-1}\!\left(\frac{\sqrt{2}-\sqrt{1}}{\sqrt{6}}\right) + \sin^{-1}\!\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}\right) + \cdots \quad (\text{to infinity}) \] \[ \textbf{Step 1: Identify the general term} \] \[ \sqrt{2}=\sqrt{1\cdot 2}, \quad \sqrt{6}=\sqrt{2\cdot 3}, \quad \sqrt{12}=\sqrt{3\cdot 4} \] \[ \text{Hence the general term is} \] \[ \sin^{-1}\!\left( \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}} \right) \] \[ \textbf{Step 2: Use inverse–sine subtraction identity} \] Let \[ \alpha_n = \sin^{-1}\!\left(\frac{1}{\sqrt{n}}\right) \] Using the identity \[ \sin(\alpha_n - \alpha_{n+1}) = \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}} \] we get \[ \sin^{-1}\!\left( \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}} \right) = \alpha_n - \alpha_{n+1} \] \[ \textbf{Step 3: Telescoping sum} \] \[ (\alpha_1 - \alpha_2) + (\alpha_2 - \alpha_3) + (\alpha_3 - \alpha_4) + \cdots \] \[ = \alpha_1 - \lim_{n\to\infty}\alpha_{n+1} \] \[ \textbf{Step 4: Evaluate limits} \] \[ \alpha_1 = \sin^{-1}(1) = \frac{\pi}{2} \] \[ \lim_{n\to\infty}\sin^{-1}\!\left(\frac{1}{\sqrt{n}}\right) = 0 \] \[ \textbf{Final Answer} \] \[ \boxed{\frac{\pi}{2}} \] \[ \textbf{Key Idea:} \] \[ \text{This is a classic telescoping inverse–trigonometric series.} \]