Sequences & Series
Sum of Series
Grade 11
Question:
<p>Let \(S_n = \dfrac{1}{1^3} + \dfrac{1+2}{1^3+2^3} + \dfrac{1+2+3}{1^3+2^3+3^3} + \ldots\ldots (n \text{ terms})\), where \(n = 1,2,3,4,\ldots\ldots\), then \(S_n\) is always less than:</p>
<p>1</p>
<p>2</p>
<p>3</p>
<p>4</p>
Step-by-Step Solution
Key Concept: Recognize that the r-th term has numerator = r(r+1)/2 and denominator = [r(r+1)/2]², so each term simplifies to 2/[r(r+1)]. Then use telescoping: 2/[r(r+1)] = 2(1/r - 1/(r+1)), making the sum telescope to 2(1 - 1/(n+1)) = 2n/(n+1).
<p><strong>Step 1:</strong> Identify the r-th term general form.</p><p>Numerator of r-th term: 1 + 2 + ... + r = r(r+1)/2</p><p>Denominator of r-th term: 1³ + 2³ + ... + r³ = [r(r+1)/2]² (using standard formula)</p><p><strong>Step 2:</strong> Simplify the r-th term.</p><p>$$T_r = \frac{r(r+1)/2}{[r(r+1)/2]^2} = \frac{1}{r(r+1)/2} = \frac{2}{r(r+1)}$$</p><p><strong>Step 3:</strong> Use partial fractions to enable telescoping.</p><p>$$T_r = \frac{2}{r(r+1)} = 2\left(\frac{1}{r} - \frac{1}{r+1}\right)$$</p><p><strong>Step 4:</strong> Sum the telescoping series.</p><p>$$S_n = \sum_{r=1}^{n} 2\left(\frac{1}{r} - \frac{1}{r+1}\right) = 2\left(\frac{1}{1} - \frac{1}{n+1}\right) = 2 - \frac{2}{n+1} = \frac{2n}{n+1}$$</p><p><strong>Step 5:</strong> Determine the upper bound.</p><p>Since $$S_n = \frac{2n}{n+1} = \frac{2n+2-2}{n+1} = 2 - \frac{2}{n+1} < 2$$ for all positive integers n.</p><p>∴ Answer: B (where option B is 2, or the smallest value greater than all possible $S_n$)</p>
Correct Answer: B