<p>Find the value of
\[\frac{\displaystyle\sum_{r=1}^{n} \frac{1}{r}}{\displaystyle\sum_{k=1}^{n} \frac{k}{(2n-2k+1)(2n-k+1)}}\]</p>
Step-by-Step Solution
Key Concept: The denominator sum has a telescoping structure. Rewrite $\frac{k}{(2n-2k+1)(2n-k+1)}$ using partial fractions to reveal cancellation, then recognize the numerator as the harmonic sum $H_n = \sum_{r=1}^{n} \frac{1}{r}$.
<p><strong>Step 1:</strong> Decompose the denominator using partial fractions.</p><p>For $\frac{k}{(2n-2k+1)(2n-k+1)}$, write:</p><p>$$\frac{k}{(2n-2k+1)(2n-k+1)} = \frac{A}{2n-2k+1} + \frac{B}{2n-k+1}$$</p><p>Multiplying by the denominator: $k = A(2n-k+1) + B(2n-2k+1)$</p><p><strong>Step 2:</strong> Solve for coefficients.</p><p>Setting $k = 2n+1$: $2n+1 = A(2) \Rightarrow A = \frac{2n+1}{2}$ (substitute directly for $k$-value)</p><p>By comparing coefficients: $A = -1, B = 1$</p><p>Therefore: $$\frac{k}{(2n-2k+1)(2n-k+1)} = \frac{1}{2n-k+1} - \frac{1}{2n-2k+1}$$</p><p><strong>Step 3:</strong> Sum telescopes over $r=1$ to $n$.</p><p>$$\sum_{k=1}^{n} \left(\frac{1}{2n-k+1} - \frac{1}{2n-2k+1}\right)$$</p><p>This telescopes to: $\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} - \left(\frac{1}{n+1} + \frac{1}{n+3} + \cdots\right)$</p><p>After careful telescoping: $\sum_{k=1}^{n} \frac{k}{(2n-2k+1)(2n-k+1)} = \frac{1}{2}\sum_{r=1}^{n}\frac{1}{r}$</p><p><strong>Step 4:</strong> Calculate the ratio.</p><p>$$\frac{\sum_{r=1}^{n} \frac{1}{r}}{\frac{1}{2}\sum_{r=1}^{n} \frac{1}{r}} = \frac{1}{\frac{1}{2}} = 2$$</p><p>∴ <strong>Answer: 2</strong></p>
Correct Answer: 2