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

Solution

Maths sir · 4 steps

Call 2ⁿ just m, and the sum becomes a Riemann sum for ∫₀¹ dx/√(1-x).

1. Rename the power. Put m = 2ⁿ. As n → ∞, m → ∞ too. The expression becomes 1/mΣᵣ₌₁ᵐ⁻¹ 1/√(1 - r/m).
2. Read as integral. Replace r/m by x and 1/m by dx. Here r/m runs from about 0 to about 1, so the limits are 0 and 1. Leaving out the single last term does not change the limit.

   lim(m → ∞) 1/mΣᵣ₌₁ᵐ⁻¹ 1/√(1-r/m) = ∫₀¹ dx/√(1-x)

3. Integrate. An antiderivative of (1-x)^(-1/2) is -2√(1-x). So the value is [-2√(1-x)]₀¹ = 0 - (-2) = 2. The integrand blows up at x = 1, but this improper integral still converges.
4. So. The limit equals 2, which is option (c).

The idea to keep. Spot 1/N outside and r/N inside. Then the sum turns into ∫₀¹ f(x) dx, whether N is n, 2ⁿ or anything else that goes to infinity.

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