Limits, Continuity & Differentiability
Evaluation of Limits
Grade 12
Question:
<p>If \(a_1 = 1\) and \(a_n = n(1 + a_{n-1})\) \(\forall\, n \geq 2\), and \(L = \lim_{n \to \infty}\left(1 + \dfrac{1}{a_1}\right)\left(1 + \dfrac{1}{a_2}\right)\cdots\left(1 + \dfrac{1}{a_n}\right)\), then</p>
<p>(a) \(L > e\)</p>
<p>(b) \(L < \pi\)</p>
<p>(c) \(L > \sqrt{e}\)</p>
<p>(d) \(L > \pi\)</p>
Step-by-Step Solution
Key Concept: Convert the recurrence relation aₙ = n(1 + aₙ₋₁) into a telescoping product by finding that (1 + 1/aₙ) = (n+1)/n, making the infinite product collapse to a simple limit.
<p><strong>Step 1: Manipulate the recurrence relation</strong></p><p>Given: aₙ = n(1 + aₙ₋₁) for n ≥ 2, with a₁ = 1</p><p>Rearrange: 1 + aₙ = 1 + n(1 + aₙ₋₁) = 1 + n + n·aₙ₋₁</p><p><strong>Step 2: Find the key ratio</strong></p><p>From aₙ = n(1 + aₙ₋₁), we get: 1 + aₙ = n(1 + aₙ₋₁) + 1</p><p>Therefore: (1 + 1/aₙ) = (aₙ + 1)/aₙ = [n(1 + aₙ₋₁) + 1]/[n(1 + aₙ₋₁)]</p><p>This simplifies to: <strong>(1 + 1/aₙ) = (n+1)/n</strong></p><p><strong>Step 3: Apply telescoping product</strong></p><p>∏ₙ₌₁ⁿ(1 + 1/aₖ) = (2/1)·(3/2)·(4/3)·...·((n+1)/n)</p><p>This telescopes to: (n+1)/1 = n+1</p><p><strong>Step 4: Take the limit</strong></p><p>L = lim_{n→∞} (n+1) = <strong>∞</strong></p><p>∴ Answer: C (The limit is infinity)</p>
Correct Answer: C