1 + frac{1}{3!} + frac{1}{5!} + frac{1}{7!} + | vedclass

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(D) We know the expansion of $e^x$ is $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \infty$. Setting $x = 1$,we get $e = 1 +

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

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(D) We know the expansion of $e^x$ is $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \infty$. Setting $x = 1$,we get $e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots \infty$. Setting $x = -1$,we get $e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots \infty$. Subtracting the two series: $e - e^{-1} = (1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots) - (1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \dots) = 2(1 + \frac{1}{3!} + \frac{1}{5!} + \dots)$. Therefore,$1 + \frac{1}{3!} + \frac{1}{5!} + \dots = \frac{e - e^{-1}}{2}$.

$1 + \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + \dots \infty = $

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