A) \[{{e}^{2}}\]
B) \[{{\log }_{e}}2\]
C) \[{{\log }_{e}}3-2\]
D) \[1-{{\log }_{e}}2\]
Correct Answer: D
Solution :
Since \[{{\log }_{{{y}^{n}}}}{{x}^{m}}=\frac{m}{n}{{\log }_{y}}x\]and \[{{\log }_{x}}x=1\]. \[\therefore \,\,\,S=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+....\] Also \[{{\log }_{e}}(1+x)=x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}+....\] Putting \[x=1,S=1-{{\log }_{e}}2\].
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