lim _(n → ∞) ( n²/(n² + 1)(n + 1) + n²/(n² + 4)(n + 2) + n²/(n² + 9)(n + 3) + .... + n²/(n² + n²)(n + n) ) is equal to :

Solution

Maths sir · 5 steps

Pull n³ out of the denominator, set x = r/n, and the sum turns into an integral from 0 to 1.

1. General term. The r-th term is n²/(n²+r²)(n+r), with r running from 1 to n. Take n² out of the first bracket and n out of the second, so the denominator becomes n³(1+r²/n²)(1+r/n).
2. Convert to integral. Each term is now 1/n·1/(1+r²/n²)(1+r/n). Replace r/n by x and 1/n by dx. Since r/n goes from near 0 up to 1, the limits are 0 and 1.

   ∫₀¹ dx/(1+x)(1+x²) = ∫₀¹ ((1/2)/(1+x) + (-1/2x + 1/2)/(1+x²)) dx

3. Partial fractions. Putting x = -1 gives A = 1/2. Comparing the x² terms gives B = -1/2, and comparing the constants gives C = 1/2.
4. Integrate each piece. 1/2 ln(1+x) gives 1/2 ln 2. Next, -1/4 ln(1+x²) gives -1/4 ln 2. Last, 1/2 tan⁻¹ x gives 1/2·π/4 = π/8.
5. So. The total is π/8 + 1/2 ln 2 - 1/4 ln 2 = π/8 + 1/4 logₑ 2, which is option (a).

The idea to keep. For a limit of a sum, write the general term as 1/n f(r/n). Then r/n becomes x, 1/n becomes dx, and the range of r/n gives the limits.

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