Trigonometry & Inverse Trigonometry Question - None

Question

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

Accepted Answer

Step 1: Use the identity: tan⁻¹(A) - tan⁻¹(B) = tan⁻¹((A-B)/(1+AB))

Rearranging: tan⁻¹((A-B)/(1+AB)) = tan⁻¹(A) - tan⁻¹(B)

Step 2: Notice that 1/(k(k+1)+1) = ((k+1) - k)/(1 + k(k+1))

This matches the form with A = k+1 and B = k

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

Step 4: The sum telescopes:

S_n = Σ[tan⁻¹(k+1) - tan⁻¹(k)] from k=1 to n

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

= tan⁻¹(n+1) - tan⁻¹(1)

= tan⁻¹(n+1) - π/4

Step 5: Key results:

• S_n = tan⁻¹(n+1) - π/4

• As n → ∞: S_n → π/2 - π/4 = π/4

• S_1 = tan⁻¹(2) - π/4

• S_n is strictly increasing and bounded above by π/4

∴ Answer: Multiple valid statements including convergence to π/4, formula S_n = tan⁻¹(n+1) - π/4, and monotonic increase

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