The value of
lim _(n → ∞) 1/nΣᵣ₌₀²ⁿ⁻¹ n²/(n² + 4r²) is :

Solution

Maths sir · 4 steps

Turn r/n into x and the sum becomes an integral. Read the upper limit from 2n, not from n.

1. Divide by n squared. Divide the top and bottom of each term by n². The general term becomes 1/n · 1/(1 + 4(r/n)²).
2. Find the limits. Replace r/n by x and 1/n by dx. At r = 0, x = 0. At r = 2n - 1, x = 2 - 1/n, which tends to 2. So x runs from 0 to 2.

   ∫₀² dx/(1 + 4x²) = [1/2 tan⁻¹(2x)]₀²

3. Integrate. The 1/2 comes from the chain rule, because the derivative of tan⁻¹(2x) is 2/(1 + 4x²). Putting in the limits gives 1/2 tan⁻¹(4) - 1/2 tan⁻¹(0) = 1/2 tan⁻¹(4).
4. So. The limit equals 1/2 tan⁻¹(4), which is option (b).

The idea to keep. For a limit of a sum, write the term as 1/n f(r/n). The limits of x are the first and last values of r/n, so always check where r stops.

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