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$\sum_{n=1}^{\infty} \frac{2 n}{(2 n+1) !}$ is equal to
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The correct answer is:
$\frac{1}{e}$
$\begin{aligned} & \sum_{n=1}^{\infty} \frac{2 n}{(2 n+1) !}=\sum_{n=1}^{\infty} \frac{2 n+1-1}{(2 n+1) !} \\ &=\sum_{n=1}^{\infty}\left(\frac{1}{2 n !}-\frac{1}{(2 n+1) !}\right) \\ &=\left(\frac{1}{2 !}-\frac{1}{3 !}\right)+\left(\frac{1}{4 !}-\frac{1}{5 !}\right)+\ldots \\ &=1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\ldots \\ &=e^{-1}=\frac{1}{e}\end{aligned}$
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