Question

The sum of the series
$\left(\frac{a-b}{a}\right)+\frac{1}{2}{\left(\frac{a-b}{a}\right)}^2+\frac{1}{3}{\left(\frac{a-b}{a}\right)}^3+....$ is

• A. ${\log }_e\left(\frac{a}{b}\right)$
• B. ${\log }_e\left(\frac{a-b}{a}\right)$
• C. ${\log }_e\left(\frac{b}{a}\right)$
• D. None of these

Answer 1

Answer: (A)

${\log }_e\left(\frac{a}{b}\right)$

We have,
$\frac{a-b}{a}+\frac{1}{2}{\left(\frac{a-b}{a}\right)}^2+\frac{1}{3}{\left(\frac{a-b}{a}\right)}^3+...\infty$
$=-\log (1-\frac{a-b}{a})[\because \log (1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-...]$
$=-{\log }_e\left(\frac{b}{a}\right)={\log }_e\left(\frac{a}{b}\right)$