Definite Integration
Limit as Definite Integral
Grade 12
Question:
<p>The value of \(\displaystyle\lim_{n\to\infty}\left(\ln\!\left(\sqrt[n]{\frac{4}{n^2}}\right)+\ln\!\left(\sqrt[n]{\frac{16}{n^2}}\right)+\ln\!\left(\sqrt[n]{\frac{36}{n^2}}\right)+\cdots+\ln\!\left(\sqrt[n]{\frac{4n^2}{n^2}}\right)\right)\) equals:</p>
<p>(a) \(4\ln(2)\)</p>
<p>(b) \(2\ln(2)-2\)</p>
<p>(c) \(2\ln(2)-4\ln(4)-4\)</p>
<p>(d) \(2\ln(4)-2\)</p>
Step-by-Step Solution
Key Concept: Recognize this sum as a Riemann sum by factoring out 1/n from the logarithm arguments and rewriting as ∑ln(√[n]{4k²/n²}) = (1/n)∑ln(2k/n). The limit converts to the integral ∫₀¹ ln(2x)dx using the definition of Riemann sums.
<p><strong>Step 1:</strong> Simplify the general term. The k-th term is ln(√[n]{4k²/n²}) = (1/n)ln(4k²/n²) = (1/n)[ln(4k²) - ln(n²)] = (1/n)[2ln(2k) - 2ln(n)]</p><p><strong>Step 2:</strong> Recognize the sum structure. The sum becomes:</p><p>S_n = ∑_{k=1}^{n} (1/n)ln(4k²/n²) = (1/n)∑_{k=1}^{n} [2ln(2k) - 2ln(n)] = (2/n)∑_{k=1}^{n} ln(2k/n)</p><p><strong>Step 3:</strong> Convert to Riemann sum. Let x_k = k/n. Then:</p><p>S_n = (2/n)∑_{k=1}^{n} ln(2·k/n) = 2∑_{k=1}^{n} ln(2k/n)·(1/n) → 2∫₀¹ ln(2x)dx as n→∞</p><p><strong>Step 4:</strong> Evaluate the integral. ∫₀¹ ln(2x)dx = [x·ln(2x) - x]₀¹ = (1·ln2 - 1) - lim_{x→0⁺}(x·ln(2x) - x) = ln2 - 1 - 0 = ln2 - 1</p><p><strong>Step 5:</strong> Final answer: 2(ln2 - 1) = 2ln2 - 2 = ln4 - 2</p><p>∴ Answer: D</p>
Correct Answer: D