A) \[(x-1)-\frac{{{(x-1)}^{2}}}{2}+\frac{{{(x-1)}^{3}}}{3}-....\infty \]
B) 0
C) 1
D) none of the above
Correct Answer: A
Solution :
Key Idea: The series of \[log\,(1+x)\] is \[x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{2}}}{3}-\frac{{{x}^{4}}}{4}+....\] Now, \[{{\log }_{e}}x={{\log }_{e}}(1+(x-1))\] \[=(x-1)-\frac{{{(x-1)}^{2}}}{2}+\frac{{{(x-1)}^{3}}}{3}-....\] Note: the series of \[\log \,(1+x)\]is defined only when \[-1<x<1.\]
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