Question:

Step 1: Define the general term of the series. Let us denote the general term as: $$T_\omega = \sin^{-1}\left[\frac{2\omega+1}{\omega(\omega+1)(\sqrt{\omega^2+2\omega}+\sqrt{\omega^2-1})}\right]$$ We need to find $\displaystyle\sum_{\omega=1}^{\infty} T_\omega$. Step 2: Identify useful algebraic identities. Observe the following key identities: $$\omega^2 + 2\omega = (\omega+1)^2 - 1$$ $$\omega^2 - 1 = (\omega-1)(\omega+1)$$ These will help us simplify the expression inside the inverse sine function. Step 3: Rationalize the denominator. Multiply the argument of $\sin^{-1}$ by the conjugate expression: $$\frac{2\omega+1}{\omega(\omega+1)(\sqrt{\omega^2+2\omega}+\sqrt{\omega^2-1})} \cdot \frac{\sqrt{\omega^2+2\omega}-\sqrt{\omega^2-1}}{\sqrt{\omega^2+2\omega}-\sqrt{\omega^2-1}}$$ Step 4: Simplify the numerator and denominator after rationalization. The denominator becomes: $$\omega(\omega+1)[(\omega^2+2\omega)-(\omega^2-1)] = \omega(\omega+1)(2\omega+1)$$ The numerator becomes: $$(2\omega+1)(\sqrt{\omega^2+2\omega}-\sqrt{\omega^2-1})$$ Therefore: $$\frac{(2\omega+1)(\sqrt{\omega^2+2\omega}-\sqrt{\omega^2-1})}{\omega(\omega+1)(2\omega+1)} = \frac{\sqrt{\omega^2+2\omega}-\sqrt{\omega^2-1}}{\omega(\omega+1)}$$ Step 5: Rewrite using the algebraic identities from Step 2. Substitute $\omega^2 + 2\omega = (\omega+1)^2 - 1$: $$\frac{\sqrt{(\omega+1)^2-1}-\sqrt{\omega^2-1}}{\omega(\omega+1)}$$ Step 6: Recognize the telescoping form. This expression can be written as a difference of inverse sine terms: $$T_\omega = \sin^{-1}\left(\frac{\sqrt{(\omega+1)^2-1}}{\omega+1}\right) - \sin^{-1}\left(\frac{\sqrt{\omega^2-1}}{\omega}\right)$$ This is a telescoping series where consecutive terms cancel. Step 7: Evaluate the telescoping sum. When we sum from $\omega = 1$ to $\infty$: $$\sum_{\omega=1}^{\infty} T_\omega = \lim_{n \to \infty} \left[\sin^{-1}\left(\frac{\sqrt{n^2+2n}}{n+1}\right) - \sin^{-1}\left(\frac{\sqrt{1-1}}{1}\right)\right]$$ As $\omega \to \infty$: $\sin^{-1}\left(\frac{\sqrt{\omega^2+2\omega}}{\omega+1}\right) \to \sin^{-1}(1) = \frac{\pi}{2}$ At $\omega = 1$: $\sin^{-1}\left(\frac{\sqrt{0}}{1}\right) = \sin^{-1}(0) = 0$ Therefore: $$\sum_{\omega=1}^{\infty} T_\omega = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$ **Final Answer:** The value of the infinite series is $\boxed{\dfrac{\pi}{2}}$, which corresponds to **Option 4**.