A) \[1/e\]
B) \[e\]
C) \[2\,e\]
D) \[3e\]
Correct Answer: B
Solution :
\[\frac{2}{1!}+\frac{4}{3!}+\frac{6}{5!}+\frac{8}{7!}+....\infty \] \[=\frac{(1+1)}{1!}+\frac{(1+3)}{3!}+\frac{(5+1)}{5!}+\frac{(7+1)}{7!}+....\infty \] \[=\left( \frac{1}{1!}+\frac{1}{3!}+\frac{1}{5!}+\frac{1}{7!}+....\infty \right)+\left( 1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+....\infty \right)\] \[=\frac{e-{{e}^{-1}}}{2}+\frac{e+{{e}^{-1}}}{2}=e\].
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