A) \[e\]
B) \[2\,e\]
C) \[{{e}^{2}}\]
D) \[1/e\]
Correct Answer: D
Solution :
\[S=\frac{2}{3\ !}+\frac{4}{5\ !}+\frac{6}{7\ !}+......+\frac{2n}{(2n+1)\ !}+......\] Here \[{{T}_{n}}=\frac{(2n+1)-1}{(2n+1)\ !}=\frac{1}{(2n)\ !}-\frac{1}{(2n+1)\ !}\] \[\Rightarrow S=\sum\limits_{n=1}^{\infty }{{{T}_{n}}}=\left( \frac{1}{2\ !}+\frac{1}{4\ !}+\frac{1}{6\ !}+... \right)\]\[-\left( \frac{1}{3\ !}+\frac{1}{5\ !}+\frac{1}{7\ !}+..... \right)\] \[=\left( \frac{e+{{e}^{-1}}}{2}-1 \right)-\left( \frac{e-{{e}^{-1}}}{2}-1 \right)={{e}^{-1}}=\frac{1}{e}\].
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