<p>The sum of the series \(\dfrac{1}{1 \cdot 2} - \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} - \cdots\) up to \(\infty\) is equal to</p>
Step-by-Step Solution
Key Concept: Use partial fractions to decompose 1/(n(n+1)) = 1/n - 1/(n+1), then recognize the alternating series forms two telescoping subseries that can be recombined using the Dirichlet eta function or by pairing consecutive terms strategically.
<p><strong>Step 1: Decompose using partial fractions</strong></p><p>For each term: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$</p><p><strong>Step 2: Rewrite the series</strong></p><p>$$S = \left(\frac{1}{1} - \frac{1}{2}\right) - \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) - \left(\frac{1}{4} - \frac{1}{5}\right) + \cdots$$</p><p><strong>Step 3: Expand and reorganize</strong></p><p>$$S = \frac{1}{1} - \frac{1}{2} - \frac{1}{2} + \frac{1}{3} + \frac{1}{3} - \frac{1}{4} - \frac{1}{4} + \frac{1}{5} + \cdots$$</p><p>Grouping terms with same denominator:</p><p>$$S = 1 - 2\left(\frac{1}{2}\right) + 2\left(\frac{1}{3}\right) - 2\left(\frac{1}{4}\right) + 2\left(\frac{1}{5}\right) - \cdots$$</p><p><strong>Step 4: Factor and recognize pattern</strong></p><p>$$S = 1 + 2\left(-\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots\right)$$</p><p>The series in parentheses equals $\ln(2) - 1$ (alternating harmonic series minus 1)</p><p>$$S = 1 + 2(\ln 2 - 1) = 2\ln 2 - 1$$</p><p>∴ Answer: D</p>
Correct Answer: D