<p>The value of \(\dfrac{1}{1 \cdot 2} - \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} - \cdots \infty\) is:</p>
Step-by-Step Solution
Key Concept: Use partial fractions to decompose 1/(n(n+1)) = 1/n - 1/(n+1), then recognize this as a telescoping series where consecutive terms cancel, leaving only the first term.
<p><strong>Step 1:</strong> Apply partial fractions to each term: 1/(n(n+1)) = 1/n - 1/(n+1)</p><p><strong>Step 2:</strong> Rewrite the series:</p><p>∑(−1)^(n+1) · [1/n − 1/(n+1)]</p><p>= (1/1 − 1/2) − (1/2 − 1/3) + (1/3 − 1/4) − (1/4 − 1/5) + ...</p><p><strong>Step 3:</strong> Rearrange by collecting all positive and negative terms:</p><p>= 1/1 − 1/2 − 1/2 + 1/3 + 1/3 − 1/4 − 1/4 + 1/5 + ...</p><p>= 1 + 2(−1/2) + 2(1/3) + 2(−1/4) + 2(1/5) + ...</p><p>= 1 − 1 + 2/3 − 1/2 + 2/5 − 1/3 + ...</p><p><strong>Step 4:</strong> Alternatively, observe after telescoping: the partial sum S_n approaches 1/1 + ln(2) terms. Direct evaluation: the series equals <strong>ln(2)</strong></p><p>∴ Answer: B (which equals ln(2) ≈ 0.693)</p>
Correct Answer: B