• # question_answer ${{\log }_{e}}(1+x)=\sum\limits_{i=1}^{\infty }{\left[ \frac{{{(-1)}^{i+1}}{{x}^{i}}}{i} \right]}$ is defined for  [Roorkee 1990] A) $x\in (-1,\,1)$ B) Any positive (+) real x C) $x\in (-1,\,1]$ D) Any positive (+) real $x(x\ne 1)$

Correct Answer: C

Solution :

${{\log }_{e}}(1+x)=\sum\limits_{i=1}^{\infty }{\frac{{{(-1)}^{i+1}}{{x}^{i}}}{i}}$is defined for $x\in (-1,1]$ Because ${{\log }_{e}}(1+x)=x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}+....\infty$ is defined for $(-1<x\le 1)$.

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