<p>The value of \[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{\pi}{n} \cdot \frac{1}{\sin\left( \dfrac{(n+r)\pi}{4n} \right)} \] is equal to:</p>
<p>(a) \( 2\ln(\sqrt{2} - 1) \)</p>
<p>(b) \( 4\ln(\sqrt{2} - 1) \)</p>
<p>(c) \( 4\ln(\sqrt{2} + 1) \)</p>
<p>(d) \( \ln\sqrt{2} \)</p>
Step-by-Step Solution
Key Concept: Recognize this Riemann sum by substituting k/n = x and converting ∑(1/n)·f(k/n) into ∫₀¹ f(x)dx, where the argument of sine becomes (1+x)π/4 as n→∞.
<p><strong>Step 1: Identify the Riemann Sum Structure</strong></p><p>Rewrite the sum as: ∑_{r=1}^{n} (π/n) · 1/sin((n+r)π/4n)</p><p>Let k = n+r, so r = 1 gives k = n+1 and r = n gives k = 2n.</p><p>This becomes: (π/n)∑_{k=n+1}^{2n} 1/sin(kπ/4n)</p><p><strong>Step 2: Convert to Riemann Sum Form</strong></p><p>Substitute x = k/n, so k = nx and Δx = 1/n. When k goes from n+1 to 2n, x goes from (n+1)/n to 2.</p><p>As n→∞: (n+1)/n → 1, so the sum becomes:</p><p>∫₁² π/sin(πx/4) dx</p><p><strong>Step 3: Evaluate the Integral</strong></p><p>∫ 1/sin(πx/4) dx = -(4/π)ln|csc(πx/4) + cot(πx/4)| + C</p><p><strong>Step 4: Apply Limits</strong></p><p>Evaluate from x = 1 to x = 2:</p><p>At x = 2: sin(π/2) = 1, csc(π/2) = 1, cot(π/2) = 0 → ln|1| = 0</p><p>At x = 1: sin(π/4) = 1/√2, csc(π/4) = √2, cot(π/4) = 1 → ln|√2 + 1|</p><p>∴ Answer: π·(4/π)·ln(√2 + 1) = <strong>4ln(√2 + 1)</strong> or equivalently <strong>2ln(2 + √2)</strong></p>
Correct Answer: C