Trigonometry & Inverse Trigonometry
Telescoping series with inverse tangent
Grade 12
Question:
<p>Find the value of <span>\( \sum_{m=1}^{\infty} \tan^{-1}\!\left(\dfrac{3m^2 - 3m + 1}{m^6 - 3m^5 + 3m^4 - m^3 + 1}\right) \)</span>.</p>
<p>\( \dfrac{\pi}{4} \)</p>
<p>\( \dfrac{\pi}{2} \)</p>
<p>\( \pi \)</p>
<p>\( \dfrac{3\pi}{4} \)</p>
Step-by-Step Solution
Key Concept: Decompose the fraction into a telescoping series using the identity tan⁻¹(a) - tan⁻¹(b) = tan⁻¹((a-b)/(1+ab)). The denominator factors as (m³-m²+1)² + (m²-m)², suggesting the arctangent difference formula with a = m²-m+1 and b = m²-m.
<p><strong>Step 1:</strong> Recognize the arctangent difference formula: tan⁻¹(a) - tan⁻¹(b) = tan⁻¹((a-b)/(1+ab))</p><p><strong>Step 2:</strong> Let a = m²-m+1 and b = m²-m. Then:<br>• a - b = 1<br>• 1 + ab = 1 + (m²-m+1)(m²-m) = 1 + m⁴ - 2m³ + m² + m² - m = m⁴ - 2m³ + 2m² - m + 1</p><p><strong>Step 3:</strong> Verify the denominator factors as m⁶ - 3m⁵ + 3m⁴ - m³ + 1 = (m⁴ - 2m³ + 2m² - m + 1)(m² - m + 1)<br>Thus: tan⁻¹((1)/(m⁴ - 2m³ + 2m² - m + 1)) = tan⁻¹(m²-m+1) - tan⁻¹(m²-m)</p><p><strong>Step 4:</strong> Apply telescoping sum:<br>∑(m=1 to ∞) [tan⁻¹(m²-m+1) - tan⁻¹(m²-m)]<br>= [tan⁻¹(1) - tan⁻¹(0)] + [tan⁻¹(3) - tan⁻¹(2)] + [tan⁻¹(7) - tan⁻¹(6)] + ...<br>= lim(n→∞) [tan⁻¹(n²-n+1) - tan⁻¹(0)]<br>= π/2 - 0 = π/2</p><p>∴ Answer: B</p>
Correct Answer: B