lim _(n → ∞) tan { Σᵣ₌₁ⁿ tan ⁻¹( 1/(1 + r + r²) ) } is equal to ______.

Solution

Maths sir · 4 steps

Write the 1 on top as (r+1)-r, and each term splits into a difference of two inverse tangents.

1. Spot the pattern. The denominator 1 + r + r² equals 1 + r(r+1), and the numerator 1 equals (r+1) - r. This is exactly the form (x-y)/(1+xy).

   tan⁻¹(1/(1+r+r²)) = tan⁻¹(((r+1)-r)/(1+r(r+1))) = tan⁻¹(r+1) - tan⁻¹ r

2. Telescope the sum. Adding from r = 1 to n, every middle term cancels with its neighbour. What remains is tan⁻¹(n+1) - tan⁻¹ 1 = tan⁻¹(n+1) - π/4.
3. Take the limit. As n → ∞, tan⁻¹(n+1) → π/2. So the sum tends to π/2 - π/4 = π/4.
4. So. The required value is tan π/4 = 1, so the answer is 1.

The common slip: 0.785. Stopping at the value of the sum, π/4, gives 0.785. The question asks for the tangent of that sum, which is 1.

The idea to keep. Whenever you see tan⁻¹ of 1/(1+product), try to write the 1 as a difference of the two factors. Then the series telescopes to its first and last terms.

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