Question:

Step 1: Simplify the denominator by recognizing a key algebraic identity. We need to verify that the denominator can be written as a product plus 1. Let's check: $$m^6 - 3m^5 + 3m^4 - m^3 + 1 = m^3(m-1)^3 + 1$$ Expanding the right side: $m^3(m-1)^3 = m^3(m^3 - 3m^2 + 3m - 1) = m^6 - 3m^5 + 3m^4 - m^3$ Therefore: $m^3(m-1)^3 + 1 = m^6 - 3m^5 + 3m^4 - m^3 + 1$ ✓ Step 2: Simplify the numerator by recognizing another algebraic identity. We claim that: $$3m^2 - 3m + 1 = m^3 - (m-1)^3$$ Expanding the right side: $$(m-1)^3 = m^3 - 3m^2 + 3m - 1$$ Therefore: $$m^3 - (m-1)^3 = m^3 - (m^3 - 3m^2 + 3m - 1) = 3m^2 - 3m + 1$$ ✓ Step 3: Apply the inverse tangent subtraction formula. Recall the identity: $\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\dfrac{a-b}{1+ab}\right)$ Let $a = m^3$ and $b = (m-1)^3$. Then: $$\tan^{-1}\left(\dfrac{m^3 - (m-1)^3}{1 + m^3(m-1)^3}\right) = \tan^{-1}(m^3) - \tan^{-1}((m-1)^3)$$ Step 4: Recognize the telescoping series structure. The original sum becomes: $$S = \sum_{m=1}^{\infty} \left[\tan^{-1}(m^3) - \tan^{-1}((m-1)^3)\right]$$ Writing out the first few terms: $$S = \left[\tan^{-1}(1^3) - \tan^{-1}(0^3)\right] + \left[\tan^{-1}(2^3) - \tan^{-1}(1^3)\right] + \left[\tan^{-1}(3^3) - \tan^{-1}(2^3)\right] + \cdots$$ Step 5: Evaluate the telescoping sum. Most terms cancel, leaving only: $$S = \lim_{N \to \infty} \tan^{-1}(N^3) - \tan^{-1}(0)$$ As $N \to \infty$, we have $\tan^{-1}(N^3) \to \dfrac{\pi}{2}$ Also, $\tan^{-1}(0) = 0$ Therefore: $$S = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$ **Final Answer:** The value of the infinite series is $\boxed{\dfrac{\pi}{2}}$, which corresponds to **Option 1**.