Question

The value of $-\frac{1}{2}-\frac{x}{3}-\frac{x^2}{4}-\frac{x^3}{5}-\dots \dots .\infty =$

• A. $\frac{{\log }_e(1-x)+x}{x^2}$
• B. $\frac{{\log }_e(1-x)-x}{x^2}$
• C. $\frac{{\log }_e(1+x)-x}{x^2}$
• D. $\frac{{\log }_e(1+x)+x}{x^2}$

Answer 1

Answer: (A)

$\frac{{\log }_e(1-x)+x}{x^2}$

As all the terms in the problem statement are negative, we should choose the expansion for $\log (1-x)$.
Thus,
$\log (1-x)=-(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+....)$
$\left(\log \right(1-x)+x)=-(\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+....)$
$\frac{\log (1-x)+x}{x^2}=-\frac{1}{2}-\frac{x}{3}-\frac{x^2}{4}-\frac{x^3}{5}-....\infty$