<p>If \(\lim_{x \to 0} \left(\sum_{r=1}^n \cos\dfrac{r\pi}{2n}\right)\left(\sum_{r=1}^n \cos^2\dfrac{r\pi}{2n}\right)\left(\sum_{r=1}^n \cos^3\dfrac{r\pi}{2n}\right)\left(\sum_{r=1}^n \cos^4\dfrac{r\pi}{2n}\right)\left(\dfrac{1}{e^n - 1}\right)^4 = \dfrac{k}{\pi^2}\), then find the value of \(\dfrac{1}{k}\).</p>
Step-by-Step Solution
Key Concept: Recognize that each sum can be approximated as a Riemann sum as n→∞, converting the sums to definite integrals. The factor (1/(e^n-1))^4 approaches 0, so we need to analyze the limiting behavior carefully by scaling the argument appropriately.
<p><strong>Step 1: Interpret the limit structure.</strong> The limit appears to be as n→∞ (the variable that actually matters). Each sum ∑_{r=1}^n cos^k(rπ/2n) is a Riemann sum for ∫₀¹ cos^k(πt) dt with partition width π/2n.</p><p><strong>Step 2: Convert sums to integrals.</strong> Let ∑_{r=1}^n cos(rπ/2n) ≈ (2n/π)∫₀^(π/2) cos(u) du = (2n/π)·1 = 2n/π</p><p>∑_{r=1}^n cos²(rπ/2n) ≈ (2n/π)∫₀^(π/2) cos²(u) du = (2n/π)·(π/4) = n/2</p><p>∑_{r=1}^n cos³(rπ/2n) ≈ (2n/π)∫₀^(π/2) cos³(u) du = (2n/π)·(2/3) = 4n/3π</p><p>∑_{r=1}^n cos⁴(rπ/2n) ≈ (2n/π)∫₀^(π/2) cos⁴(u) du = (2n/π)·(3π/16) = 3n/8</p><p><strong>Step 3: Form the product.</strong> The product of the four sums is approximately: (2n/π)·(n/2)·(4n/3π)·(3n/8) = (2n·n·4n·3n)/(π·2·3π·8) = 12n⁴/(48π²) = n⁴/(4π²)</p><p><strong>Step 4: Apply the scaling factor.</strong> We have (1/(e^n-1))⁴. As n→∞, e^n dominates, so (1/(e^n-1))⁴ ≈ 1/e^(4n).</p><p><strong>Step 5: Reconsider the problem structure.</strong> For the limit to be finite and equal to k/π², we need the exponent in the denominator to match the polynomial growth. If instead the factor is (1/(e·n-1))⁴ or similar, it would give (1/n)⁴. Then: [n⁴/(4π²)]·(1/n⁴) = 1/(4π²), giving k = 4.</p><p><strong>Step 6: Calculate 1/k.</strong> If k = 4, then 1/k = 1/4.</p><p><strong>∴ Answer: 1/k = 1/4</strong></p>
Correct Answer: 1