sum_{n=1}^{infty} frac{2n}{(2n+1)!} is equal | vedclass

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Question

(A) We have the series $\sum_{n=1}^{\infty} \frac{2n}{(2n+1)!}$.Rewrite the numerator as $(2n+1) - 1$:$\sum_{n=1}^{\infty} \frac{2n+1-1}{(2n+1)!} = \s

Vedclass · Accepted Answer

$\frac{1}{e}$

Vedclass · Answer

(A) We have the series $\sum_{n=1}^{\infty} \frac{2n}{(2n+1)!}$.Rewrite the numerator as $(2n+1) - 1$:$\sum_{n=1}^{\infty} \frac{2n+1-1}{(2n+1)!} = \sum_{n=1}^{\infty} \left( \frac{2n+1}{(2n+1)!} - \frac{1}{(2n+1)!} \right) = \sum_{n=1}^{\infty} \left( \frac{1}{(2n)!} - \frac{1}{(2n+1)!} \right)$.Expanding the summation:$= \left( \frac{1}{2!} - \frac{1}{3!} \right) + \left( \frac{1}{4!} - \frac{1}{5!} \right) + \dots$Recall the Taylor series for $e^x$ at $x=1$: $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$ and $e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$Thus,$\frac{1}{e} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots$Therefore,the sum is equal to $\frac{1}{e}$.

$\sum_{n=1}^{\infty} \frac{2n}{(2n+1)!}$ is equal to

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