<p>The sum of the series \(\dfrac{1}{2!} - \dfrac{1}{3!} + \dfrac{1}{4!} - \cdots\) upto infinity is</p>
Step-by-Step Solution
Key Concept: Recognize that this is an alternating series related to the Taylor expansion of e^x. The series ∑((-1)^(n+1))/(n+1)! can be connected to e^x = ∑(x^n)/n! by separating terms or using the fact that e^(-1) = ∑(-1)^n/n!.
<p><strong>Step 1:</strong> Recall that <em>e</em> = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ...</p><p><strong>Step 2:</strong> Also, 1/<em>e</em> = <em>e</em>^(-1) = 1 - 1/1! + 1/2! - 1/3! + 1/4! - ...</p><p><strong>Step 3:</strong> The given series is: 1/2! - 1/3! + 1/4! - ... = (1/2! - 1/3! + 1/4! - ...) </p><p><strong>Step 4:</strong> Rewrite 1/<em>e</em> = 1 - 1 + 1/2! - 1/3! + 1/4! - ...</p><p><strong>Step 5:</strong> Therefore: 1/2! - 1/3! + 1/4! - ... = 1/<em>e</em> - (1 - 1) = 1/<em>e</em> - 1</p><p><strong>Step 6:</strong> Simplifying: Sum = 1/<em>e</em> - 1 = (1 - <em>e</em>)/<em>e</em></p><p>∴ Answer: <strong>B</strong> (which is 1/<em>e</em> - 1 or equivalent form)</p>
Correct Answer: B