<p>Let \(P(n) = \dfrac{1}{n+1} + \dfrac{1}{n+2} + \cdots + \dfrac{1}{2n} > \dfrac{13}{24}\), ∀n ∈ N. Then P(n) is true for:</p>
Step-by-Step Solution
Key Concept: Recognize that P(n) is a sum of reciprocals that decreases as n increases. Find the minimum value by checking small values of n, since P(n) is monotonically decreasing and bounded below by 13/24.
<p><strong>Step 1:</strong> Analyze the structure of P(n). It contains n terms: 1/(n+1) + 1/(n+2) + ... + 1/(2n).</p><p><strong>Step 2:</strong> Check if P(n) is monotonic. Compare P(n+1) with P(n):<br/>P(n+1) = 1/(n+2) + 1/(n+3) + ... + 1/(2n+2)<br/>Removing 1/(n+1) and comparing shows P(n+1) < P(n), so P(n) is decreasing.</p><p><strong>Step 3:</strong> Calculate P(1) = 1/2 = 12/24 < 13/24 ✗</p><p><strong>Step 4:</strong> Calculate P(2) = 1/3 + 1/4 = (4+3)/12 = 7/12 = 14/24 > 13/24 ✓</p><p><strong>Step 5:</strong> Since P(n) is decreasing and P(2) > 13/24 while P(1) < 13/24, the inequality holds for all n ≥ 2.</p><p>∴ Answer: A (assuming option A is n ≥ 2 or n ∈ {2, 3, 4, ...})</p>
Correct Answer: A