Sequences & Series
Telescoping series and limits
Grade 11
Question:
<p>If \(S_n = \displaystyle\sum_{r=1}^{n} \dfrac{6r+9}{(r+1)^2(r+2)^2}\) and \(\lim_{n \to \infty} S_n = \dfrac{p}{q}\) (in lowest terms), find \(|p - q|_{\text{least}}\).</p>
Step-by-Step Solution
Key Concept: Decompose the fraction using partial fractions to recognize a telescoping series: 6r+9 = 3(r+1)² + 3(r+2)² - 6(r+1)(r+2), allowing terms to cancel and revealing that only boundary terms survive.
<p><strong>Step 1: Partial Fraction Decomposition</strong></p><p>Rewrite the numerator strategically:</p><p>6r + 9 = 3[(r+1)² + (r+2)² - 2(r+1)(r+2)]</p><p>This suggests decomposing:</p><p>$$\frac{6r+9}{(r+1)^2(r+2)^2} = \frac{3}{(r+1)^2} - \frac{3}{(r+2)^2}$$</p><p><strong>Step 2: Verify the decomposition</strong></p><p>RHS = $$\frac{3(r+2)^2 - 3(r+1)^2}{(r+1)^2(r+2)^2} = \frac{3[(r+2)^2-(r+1)^2]}{(r+1)^2(r+2)^2}$$</p><p>= $$\frac{3[(r^2+4r+4)-(r^2+2r+1)]}{(r+1)^2(r+2)^2} = \frac{3(2r+3)}{(r+1)^2(r+2)^2}$$ ✓</p><p><strong>Step 3: Sum the telescoping series</strong></p><p>$$S_n = \sum_{r=1}^{n}\left[\frac{3}{(r+1)^2} - \frac{3}{(r+2)^2}\right]$$</p><p>This telescopes:</p><p>$$S_n = \frac{3}{2^2} - \frac{3}{(n+2)^2} = \frac{3}{4} - \frac{3}{(n+2)^2}$$</p><p><strong>Step 4: Take the limit</strong></p><p>$$\lim_{n \to \infty} S_n = \frac{3}{4} - 0 = \frac{3}{4}$$</p><p>So p = 3, q = 4 (in lowest terms)</p><p><strong>Step 5: Find |p - q|</strong></p><p>|p - q| = |3 - 4| = <strong>1</strong></p>
Correct Answer: 1