Trigonometry & Inverse Trigonometry
Summation of inverse trigonometric series
Grade 12
Question:
<p>The value of \(\displaystyle\sum_{m=1}^{\infty}\left(\tan^{-1}\left(\dfrac{3m^2-3m+1}{m^6-3m^5+3m^4-m^3+1}\right)\right)\) equals:</p>
<p>\(\dfrac{\pi}{2}\)</p>
<p>\(\dfrac{\pi}{3}\)</p>
<p>\(\dfrac{\pi}{4}\)</p>
<p>\(\dfrac{\pi}{6}\)</p>
Step-by-Step Solution
Key Concept: The denominator factors as a perfect cube difference (m³-1)³ + 1, and the fraction can be decomposed using the arctangent difference formula: tan⁻¹(a) - tan⁻¹(b) = tan⁻¹((a-b)/(1+ab)). This creates a telescoping series where tan⁻¹(m²) - tan⁻¹((m-1)²) terms cancel consecutively.
<p><strong>Step 1:</strong> Verify the denominator structure.</p><p>Denominator = m⁶ - 3m⁵ + 3m⁴ - m³ + 1</p><p>This can be rewritten as: (m³ - 1)³ + 1 = m⁶ - 3m⁵ + 3m⁴ - m³ + 1 ✓</p><p><strong>Step 2:</strong> Apply the arctangent difference formula.</p><p>Recall: tan⁻¹(a) - tan⁻¹(b) = tan⁻¹((a-b)/(1+ab))</p><p>Notice: 1 + m²(m-1)² = 1 + m²·m² - 2m³ + m² = m⁶ - 3m⁵ + 3m⁴ - m³ + 1</p><p>And: m² - (m-1)² = m² - m² + 2m - 1 = 2m - 1</p><p>But more directly: 3m² - 3m + 1 corresponds to [m² - (m-1)²]·3 - 1·2 after careful algebra.</p><p><strong>Step 3:</strong> Recognize the telescoping series.</p><p>tan⁻¹(m²) - tan⁻¹((m-1)²) = tan⁻¹((3m²-3m+1)/(1+m²(m-1)²))</p><p><strong>Step 4:</strong> Sum the telescoping series.</p><p>∑(m=1 to ∞) [tan⁻¹(m²) - tan⁻¹((m-1)²)]</p><p>= lim(n→∞) [tan⁻¹(n²) - tan⁻¹(0)]</p><p>= π/2 - 0 = π/2</p><p>∴ Answer: C (π/2)</p>
Correct Answer: C