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

Solution

Maths sir · 3 steps

Base goes to 1 and power goes to infinity, so write it as e raised to (base minus 1) times the power.

1. Spot the form. Let Hₙ = 1 + 1/2 + … + 1/n. Since Hₙ ≤ n, the extra term Hₙ/n² ≤ 1/n → 0. So the base tends to 1 while the power n → ∞. This is a 1^∞ form.

   lim(n → ∞)(1 + Hₙ/n²)ⁿ = e^(lim(n → ∞) n · Hₙ/n²) = e^(lim(n → ∞) Hₙ/n)

2. Find the exponent. The harmonic sum grows like a logarithm: Hₙ ≤ 1 + ln n. Since ln n grows far slower than n, Hₙ/n ≤ (1 + ln n)/n → 0. Also Hₙ/n > 0, so the exponent tends to 0.
3. So. The limit is e⁰ = 1, which is option (b).

The idea to keep. For a 1^∞ limit (1 + aₙ)^(bₙ) with aₙ → 0, the answer is e^(lim aₙ bₙ). Remember that 1 + 1/2 + … + 1/n grows only like ln n.

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