Question

$\frac{a-b}{a}+\frac{1}{2}(\frac{a-b}{a}{)}^2+\frac{1}{3}(\frac{a-b}{a}{)}^3+=\dots$

• A. ${\log }_e\left(\frac{a}{b}\right)$
• B. ${\log }_e\left(\frac{b}{a}\right)$
• C. ${\log }_e(a+b)$
• D. ${\log }_e(a-b)$

Answer 1

The correct option is A ${\log }_e\left(\frac{a}{b}\right)$

$\frac{a-b}{a}+\frac{1}{2}(\frac{a-b}{a}{)}^2+\frac{1}{3}(\frac{a-b}{a}{)}^3+=\dots$
We know $\log (1-X)=-\sum \frac{X^n}{n}$
So, here $X=\left(\frac{a-b}{a}\right)$
$S=-\log \left(1-\frac{(a-b)}{a}\right)$
$=-\log \left(\frac{b}{a}\right)$
$S=\log \left(\frac{a}{b}\right)$