Question:

Step 1: For an infinite G.P. to have a finite sum, the common ratio $r$ must satisfy $|r| < 1$. Step 2: Check option (1): $\dfrac{1}{\sin^2\theta}$. Since $0 < \sin^2\theta \leq 1$, we have $\dfrac{1}{\sin^2\theta} \geq 1$. So $|r| \geq 1$, which does NOT satisfy $|r| < 1$. Option (1) is incorrect. Step 3: Check option (2): $e^{-t^2}$ for $t \neq 0$. Since $t^2 > 0$, we have $-t^2 < 0$, so $0 < e^{-t^2} < 1$. Thus $|r| < 1$. Option (2) is correct. Step 4: Check option (3): $\dfrac{1}{2}\left(y^2 + \dfrac{1}{y^2}\right)$. By AM-GM, $y^2 + \dfrac{1}{y^2} \geq 2$, so $\dfrac{1}{2}\left(y^2 + \dfrac{1}{y^2}\right) \geq 1$. Thus $|r| \geq 1$. Option (3) is incorrect. Step 5: Check option (4): $\dfrac{2}{x^2 - 4x + 7}$. Complete the square: $x^2 - 4x + 7 = (x-2)^2 + 3 \geq 3$. So $\dfrac{2}{x^2-4x+7} \leq \dfrac{2}{3} < 1$ and $> 0$. Thus $0 < r \leq \dfrac{2}{3} < 1$. Option (4) is correct.