Trigonometry & Inverse Trigonometry Question - None

Question

Let $S_n = \displaystyle\sum_{k=1}^{n} \tan^{-1}\left(\dfrac{1}{k(k+1)+1}\right)$ for positive integers $n \in N$, then:

Accepted Answer

Step 1: Use the arctangent difference formula: tan⁻¹(a) - tan⁻¹(b) = tan⁻¹((a-b)/(1+ab))

Step 2: Verify the identity: tan⁻¹(k+1) - tan⁻¹(k) = tan⁻¹(((k+1)-k)/(1+k(k+1))) = tan⁻¹(1/(k²+k+1))

Step 3: Therefore: tan⁻¹(1/(k(k+1)+1)) = tan⁻¹(k+1) - tan⁻¹(k)

Step 4: Write out the sum:

Sₙ = [tan⁻¹(2) - tan⁻¹(1)] + [tan⁻¹(3) - tan⁻¹(2)] + ... + [tan⁻¹(n+1) - tan⁻¹(n)]

Step 5: This telescopes to: Sₙ = tan⁻¹(n+1) - tan⁻¹(1) = tan⁻¹(n+1) - π/4

Step 6: As n → ∞: Sₙ → π/2 - π/4 = π/4

∴ Answer: Sₙ = tan⁻¹(n+1) - π/4 (or equivalent statements about limits and specific values)

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