The sum of the series frac{1}{2!} - frac{1}{3 | vedclass

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(D) We know the Taylor series expansion for $e^x$ is given by:$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$Substitute $x =

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$e^{-1}$

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(D) We know the Taylor series expansion for $e^x$ is given by:$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$Substitute $x = -1$ into the expansion:$e^{-1} = 1 + (-1) + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!} + \frac{(-1)^4}{4!} + \dots$Simplifying the terms:$e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$Since $1 - 1 = 0$,we get:$e^{-1} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$Thus,the sum of the series is $e^{-1}$.

The sum of the series $\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$ up to infinity is:

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