frac{2}{1!} + frac{4}{3!} + frac{6}{5!} + fra | vedclass

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(B) The given series is $S = \frac{2}{1!} + \frac{4}{3!} + \frac{6}{5!} + \frac{8}{7!} + \dots \infty$. We can write the $n$-th term as $\frac{2n}{(2n

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$e$

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(B) The given series is $S = \frac{2}{1!} + \frac{4}{3!} + \frac{6}{5!} + \frac{8}{7!} + \dots \infty$. We can write the $n$-th term as $\frac{2n}{(2n-1)!}$. $S = \sum_{n=1}^{\infty} \frac{2n}{(2n-1)!} = \sum_{n=1}^{\infty} \frac{(2n-1) + 1}{(2n-1)!} = \sum_{n=1}^{\infty} \left( \frac{2n-1}{(2n-1)!} + \frac{1}{(2n-1)!} \right)$. $S = \sum_{n=1}^{\infty} \frac{1}{(2n-2)!} + \sum_{n=1}^{\infty} \frac{1}{(2n-1)!}$. Expanding the terms: $S = \left( \frac{1}{0!} + \frac{1}{2!} + \frac{1}{4!} + \dots \right) + \left( \frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \dots \right)$. Recall that $e = \sum_{n=0}^{\infty} \frac{1}{n!}$ and $e^{-1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}$. Thus,$\frac{e + e^{-1}}{2} = \frac{1}{0!} + \frac{1}{2!} + \frac{1}{4!} + \dots$ and $\frac{e - e^{-1}}{2} = \frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \dots$. $S = \frac{e + e^{-1}}{2} + \frac{e - e^{-1}}{2} = \frac{2e}{2} = e$.

$\frac{2}{1!} + \frac{4}{3!} + \frac{6}{5!} + \frac{8}{7!} + \dots \infty = $

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