The sum of the series frac{1}{2!} + frac{1}{4 | vedclass

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(B) We know the expansion of $\cosh(x)$ is given by $\frac{e^x + e^{-x}}{2} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \dots$Setting $x

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$\frac{(e - 1)^2}{2e}$

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(B) We know the expansion of $\cosh(x)$ is given by $\frac{e^x + e^{-x}}{2} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \dots$Setting $x = 1$,we get:$\frac{e^1 + e^{-1}}{2} = 1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots$$\frac{e + \frac{1}{e}}{2} = 1 + \left( \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \right)$$\frac{e^2 + 1}{2e} = 1 + \left( \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \right)$Therefore,the sum of the series is:$\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots = \frac{e^2 + 1}{2e} - 1$$= \frac{e^2 + 1 - 2e}{2e} = \frac{(e - 1)^2}{2e}$.

The sum of the series $\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots$ is

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