The value of lim n → ∞ Σk=1ⁿ n³/(n²+k²)(n²+3 k²) is :

Solution

Maths sir · 4 steps

Divide top and bottom by n⁴, put x = k/n, and the sum turns into an integral from 0 to 1.

1. Sum to integral. Divide numerator and denominator by n⁴. Each term becomes 1/n · 1/(1+(k/n)²)(1+3(k/n)²). As n → ∞, k/n → x and 1/n → dx, with x running from 0 to 1.

   L = ∫₀¹ dx/(1+x²)(1+3x²)

2. Partial fractions. Treat x² as t: 1/(1+t)(1+3t) = A/(1+t) + B/(1+3t). Putting t = -1 gives A = -1/2, and putting t = -1/3 gives B = 3/2. So the integrand is -1/2 · 1/(1+x²) + 3/2 · 1/(1+3x²).
3. Integrate. ∫₀¹ dx/(1+x²) = π/4 and ∫₀¹ dx/(1+3x²) = 1/√3 tan⁻¹√3 = π/3√3. So L = -π/8 + 3/2 · π/3√3 = -π/8 + (√3 π)/6 = (4√3-3)π/24.
4. So. Rationalise option (c): (4√3+3)(4√3-3) = 48 - 9 = 39, so 13π/8(4√3+3) = 13(4√3-3)π/(8 × 39) = (4√3-3)π/24. The answer is (c).

The idea to keep. For a limit of a sum, take out enough powers of n to leave exactly one 1/n and everything else in k/n. Then k/n → x, 1/n → dx, Σ → ∫₀¹.

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