frac{2}{3!} + frac{4}{5!} + frac{6}{7!} + dot | vedclass

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(D) Let the series be $S = \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!}$.The general term is $T_n = \frac{2n}{(2n+1)!} = \frac{(2n+1)-1}{(2n+1)!} = \frac{1}

Vedclass · Accepted Answer

$1/e$

Vedclass · Answer

(D) Let the series be $S = \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!}$.The general term is $T_n = \frac{2n}{(2n+1)!} = \frac{(2n+1)-1}{(2n+1)!} = \frac{1}{(2n)!} - \frac{1}{(2n+1)!}$.Summing from $n=1$ to $\infty$:$S = \left( \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \right) - \left( \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + \dots \right)$.We know that $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$ and $e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots$.Thus,$\frac{e+e^{-1}}{2} = 1 + \frac{1}{2!} + \frac{1}{4!} + \dots \implies \frac{1}{2!} + \frac{1}{4!} + \dots = \frac{e+e^{-1}}{2} - 1$.And $\frac{e-e^{-1}}{2} = \frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \dots \implies \frac{1}{3!} + \frac{1}{5!} + \dots = \frac{e-e^{-1}}{2} - 1$.Substituting these values:$S = \left( \frac{e+e^{-1}}{2} - 1 \right) - \left( \frac{e-e^{-1}}{2} - 1 \right) = \frac{e+e^{-1}-e+e^{-1}}{2} = e^{-1} = \frac{1}{e}$.

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

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