The value of
lim _(n → ∞) 6 tan { Σᵣ₌₁ⁿ tan ⁻¹( 1/(r² + 3r + 3) ) } is equal to :

Solution

Maths sir · 4 steps

Write the denominator as 1 + (r+1)(r+2), and every term splits into a difference of two inverse tangents.

1. Spot the form. Note that r² + 3r + 3 = 1 + (r+1)(r+2), and the numerator 1 = (r+2) - (r+1). So each term has the shape tan⁻¹(a-b)/(1+ab) with a = r+2 and b = r+1.

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

2. Telescope the sum. Adding from r = 1 to n, all the middle terms cancel. Only tan⁻¹(n+2) - tan⁻¹ 2 is left.
3. Take the limit. As n → ∞, tan⁻¹(n+2) → π/2. So the sum tends to π/2 - tan⁻¹ 2 = cot⁻¹ 2 = tan⁻¹ 1/2.
4. So. The tangent of the sum tends to 1/2, and the limit is 6 × 1/2 = 3, which is option (c).

The idea to keep. When a tan⁻¹ series has a quadratic in the denominator, try to write it as 1 + ab with a - b equal to the numerator. Then the series telescopes.

Answer from the NTA final answer key. Solution by PrepWiser.