lim _(n → ∞) [ 1/(1 + n) + 1/(2 + n) + 1/(3 + n) + ... + 1/2n ] is equal to

Solution

Maths sir · 4 steps

Pull out 1/n from each term, and the sum turns into ∫₀¹ dx/(1+x).

1. Write the general term. The terms are 1/(n+r) for r = 1, 2, …, n. Divide top and bottom by n: 1/(n+r) = 1/n · 1/(1 + r/n).
2. Read off the integral. Replace r/n by x and 1/n by dx. As r runs from 1 to n, r/n runs from near 0 up to 1, so the limits are 0 and 1.

   lim(n → ∞) Σᵣ₌₁ⁿ 1/n · 1/(1 + r/n) = ∫₀¹ dx/(1+x)

3. Integrate. ∫₀¹ dx/(1+x) = [logₑ(1+x)]₀¹ = logₑ 2 - logₑ 1 = logₑ 2.
4. So. The limit equals logₑ 2, which is option (b).

The idea to keep. For a limit of a sum, write each 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.