Sum of the infinite series 

​$$S = \frac{1}{2} - \frac{1}{3 \times 1!} + \frac{1}{4 \times 2!} - \frac{1}{5 \times 3!} + \cdots$$

is equal to ?

Correct option is B

Given series is:

​$$S = \frac{1}{2} - \frac{1}{3 \times 1!} + \frac{1}{4 \times 2!} - \frac{1}{5 \times 3!} + \cdots\\[10pt]\implies S=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{(n+1)(n-1)!} \\[10pt]=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{n(n+1)(n-1)!}=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n+1-1}{(n+1)!}\\[10pt]S = \sum_{n=1}^{\infty} (-1)^{n-1} \left( \frac{1}{n!} - \frac{1}{(n+1)!} \right) \\[10pt]\text{Expanding the first few terms:} \\[10pt]S = \left( \frac{1}{1!} - \frac{1}{2!} \right) - \left( \frac{1}{2!} - \frac{1}{3!} \right) + \left( \frac{1}{3!} - \frac{1}{4!} \right) - \left( \frac{1}{4!} - \frac{1}{5!} \right) + \dots \\[10pt]\text{Rearrange the terms:} \\[10pt]S = \frac{1}{1!} - \frac{1}{2!} - \frac{1}{2!} + \frac{1}{3!} + \frac{1}{3!} - \frac{1}{4!} - \frac{1}{4!} + \frac{1}{5!} + \dots \\[10pt]\text{Grouping similar terms:} \\[10pt]S = 1 - 2 \left( \frac{1}{2!} \right) + 2 \left( \frac{1}{3!} \right) - 2 \left( \frac{1}{4!} \right) + \dots \\[10pt]\text{Recognizing the pattern from the Taylor series of } e^x: \\[10pt]e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \\[10pt]\text{For } x = -1, \text{ we get:} \\[10pt]e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots \\[10pt]\frac{1}{e} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots \\[10pt]\text{Substituting this back into } S: \\[10pt]S = 1 - 2 \left( \frac{1}{e} \right) \\[10pt]S = 1 - \frac{2}{e} \\[10pt]$$

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