A) \[15e\]
B) \[{{e}^{1/2}}+e\]
C) \[{{e}^{1/2}}-1\]
D) \[{{e}^{1/2}}-e\]
Correct Answer: C
Solution :
The nth term of given series is \[{{T}_{n}}=\frac{1.3.5.7.....(2n-1)}{1.2.3.4.........(2n)}\] \[{{T}_{n}}=\frac{1.2.3.4.....(2n-2)(2n-1)(2n)}{1.2.3.4...(2n-1)(2n)}\]\[\times \frac{1}{2.4.6...(2n-2)(2n)}\] \[{{T}_{n}}\]\[=\frac{1}{({{2}^{n}}n!)}\],\[\therefore \,\,S=\sum\limits_{n=1}^{\infty }{\frac{{{\left( \frac{1}{2} \right)}^{n}}}{n!}={{e}^{\frac{1}{2}}}-1.}\]
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