Definite Integration
Limit as Riemann Sum
Grade 12
Question:
<p>If \(\lim_{n \to \infty}\left(\dfrac{1}{\sqrt{n}\sqrt{n+1}} + \dfrac{1}{\sqrt{n}\sqrt{n+2}} + \cdots + \dfrac{1}{\sqrt{n}\sqrt{n+n}}\right)\) is equal to \(a\sqrt{b} - \dfrac{c}{d}\), where <i>a</i>, <i>b</i>, <i>c</i> and <i>d</i> are positive integers, <i>c</i> and <i>d</i> are co-prime. Find the value of \((a^4 + b^3 + c^2 + d)\).</p>
Step-by-Step Solution
Key Concept: Convert the Riemann sum to a definite integral by factoring out 1/n from the denominator and recognizing the sum as ∫₀¹ 1/√(1+x) dx. Use substitution u = 1+x to evaluate.
<p><strong>Step 1:</strong> Rewrite the sum by factoring:<br>$$\sum_{k=1}^{n} \frac{1}{\sqrt{n}\sqrt{n+k}} = \sum_{k=1}^{n} \frac{1}{n\sqrt{1+\frac{k}{n}}} = \frac{1}{n}\sum_{k=1}^{n} \frac{1}{\sqrt{1+\frac{k}{n}}}$$</p><p><strong>Step 2:</strong> Recognize this as a Riemann sum with $\Delta x = \frac{1}{n}$ and partition points $x_k = \frac{k}{n}$:<br>$$\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n} \frac{1}{\sqrt{1+\frac{k}{n}}} = \int_0^1 \frac{1}{\sqrt{1+x}} dx$$</p><p><strong>Step 3:</strong> Evaluate using substitution $u = 1+x$, so $du = dx$:<br>$$\int_0^1 \frac{1}{\sqrt{1+x}} dx = \int_1^2 u^{-1/2} du = [2u^{1/2}]_1^2 = 2\sqrt{2} - 2$$</p><p><strong>Step 4:</strong> Express as $a\sqrt{b} - \frac{c}{d}$:<br>$$2\sqrt{2} - 2 = 2\sqrt{2} - \frac{2}{1}$$<br>So $a = 2$, $b = 2$, $c = 2$, $d = 1$ (where gcd(2,1) = 1)</p><p><strong>Step 5:</strong> Calculate:<br>$$a^4 + b^3 + c^2 + d = 2^4 + 2^3 + 2^2 + 1 = 16 + 8 + 4 + 1 = 29$$</p><p>∴ Answer: <strong>29</strong></p>
Correct Answer: 29