A) 2
B) \[\frac{1}{2}\]
C) \[\log 3\]
D) None of these
Correct Answer: B
Solution :
Given series =\[1-\log 2+\frac{{{(\log 2)}^{2}}}{2!}-\frac{{{(\log 3)}^{3}}}{3!}+....\] \[={{e}^{-\log 2}}={{e}^{\log \left( \frac{1}{2} \right)}}\]\[=\frac{1}{2}\].
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