A) \[e\]
B) \[-1\]
C) \[2e\]
D) None of these
Correct Answer: B
Solution :
We have, \[\sum\limits_{n=1}^{\infty }{\frac{2n}{(2n+1)!}}=\sum\limits_{n=1}^{\infty }{\frac{2n+1-1}{(2n+1)!}}\] \[=\sum\limits_{n=1}^{\infty }{\left( \frac{1}{(2n!)}-\frac{1}{(2n+1)!} \right)}\] \[=\sum\limits_{n=1}^{\infty }{\frac{1}{(2n)!}}-\sum\limits_{n=1}^{\infty }{\frac{1}{(2n+1)!}}\] \[=\left[ \frac{e+{{e}^{-1}}}{2}-1 \right]-\left[ \frac{e-{{e}^{-1}}}{2}-1 \right]={{e}^{-1}}\]
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