Trigonometry & Inverse Trigonometry Question - None

Question

Let $ S_n = \displaystyle\sum_{r=1}^{n} \tan^{-1}\!\left(\dfrac{1}{1+(r+2)(r+1)}\right) $. Which of the following are correct?

(a) $S_n = \tan^{-1}(n+2) - \tan^{-1}(2)$

(b) $S_n = \displaystyle\sum_{r=1}^{n}\left[\tan^{-1}(r+2) - \tan^{-1}(r+1)\right]$

(c) $\lim_{n\to\infty} S_n = \dfrac{\pi}{2} - \tan^{-1}(2)$

Accepted Answer

Step 1: Decompose the argument using the tangent difference formula

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)

Note that 1+(r+2)(r+1) = (r+2)-(r+1) [the difference, not sum]

Step 2: Identify the telescoping form

Let A = r+2 and B = r+1. Then:

tan⁻¹((r+2)-(r+1)/(1+(r+2)(r+1))) = tan⁻¹(1/(1+(r+2)(r+1))) = tan⁻¹(r+2) - tan⁻¹(r+1)

Therefore each term: tan⁻¹(1/(1+(r+2)(r+1))) = tan⁻¹(r+2) - tan⁻¹(r+1)

Step 3: Sum the telescoping series

S_n = Σ[tan⁻¹(r+2) - tan⁻¹(r+1)] for r=1 to n

= [tan⁻¹(3) - tan⁻¹(2)] + [tan⁻¹(4) - tan⁻¹(3)] + ... + [tan⁻¹(n+2) - tan⁻¹(n+1)]

= tan⁻¹(n+2) - tan⁻¹(2)

Step 4: Evaluate the limit

lim(n→∞) S_n = lim(n→∞) [tan⁻¹(n+2) - tan⁻¹(2)]

= π/2 - tan⁻¹(2)

∴ Options A, B, and C are all correct

Question: