Definite Integration
Limit as Definite Integral
Grade 12
Question:
<p><strong>56.</strong> The value of \(\displaystyle\lim_{n \to \infty} \sum_{r=1}^{n} \frac{\pi}{n} \cdot \frac{1}{\sin\!\left(\dfrac{(n+r)\pi}{4n}\right)}\) 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 sum as a Riemann sum for the integral ∫₀¹ 1/sin(π(1+x)/4) dx by substituting r/n → x. The limits transform: when r=1, x=1/n→0; when r=n, x=1. Rewrite the general term to match the integrand form.
<p><strong>Step 1:</strong> Recognize the Riemann sum structure. The sum has the form Σ[π/n · f((n+r)π/(4n))]. Rewrite (n+r)π/(4n) = π(n+r)/(4n) = (π/4)(1 + r/n).</p><p><strong>Step 2:</strong> Let x = r/n, so Δx = 1/n. As n→∞, the sum becomes a definite integral:</p><p>$$\lim_{n \to \infty} \sum_{r=1}^{n} \frac{\pi}{n} \cdot \frac{1}{\sin\left(\frac{\pi(1+r/n)}{4}\right)} = \int_0^1 \frac{\pi}{\sin\left(\frac{\pi(1+x)}{4}\right)} dx$$</p><p><strong>Step 3:</strong> Substitute u = (1+x), so du = dx. When x=0, u=1; when x=1, u=2:</p><p>$$\int_1^2 \frac{\pi}{\sin(\pi u/4)} du$$</p><p><strong>Step 4:</strong> Integrate: ∫ csc(πu/4) du = -(4/π)ln|csc(πu/4) + cot(πu/4)| + C</p><p><strong>Step 5:</strong> Evaluate at bounds [1,2]:</p><p>At u=2: -(4/π)ln|csc(π/2) + cot(π/2)| = -(4/π)ln|1 + 0| = 0</p><p>At u=1: -(4/π)ln|csc(π/4) + cot(π/4)| = -(4/π)ln|√2 + 1|</p><p>Result: 0 - (-(4/π)ln(√2 + 1)) = (4/π)ln(√2 + 1) = (4/π)ln(1 + √2)</p><p>∴ Answer: C</p>
Correct Answer: C