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$ \sum \limits ^{\infty}_{n=1} \large\frac{2n}{(2n+1)\ !} =$

\[\begin {array} {1 1} (a)\;\frac{1}{e} & \quad (b)\;\frac{e}{2} \\ (c)\;e & \quad (d)\;2e \end {array}\]

1 Answer

$ (a)\;\frac{1}{e}$
answered Nov 7, 2013 by pady_1
 
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