Question

$2[\frac{m-n}{m+n}+\frac{1}{3}{\left(\frac{m-n}{m+n}\right)}^3+\frac{1}{5}{\left(\frac{m-n}{m+n}\right)}^5+...]$ is equal to

• A. ${\log }_e\left(\frac{m}{n}\right)$
• B. ${\log }_e\left(\frac{n}{m}\right)$
• C. ${\log }_{e}mn$
• D. $2{\log }_{e}mn$

Answer 1

Answer: (A)

${\log }_e\left(\frac{m}{n}\right)$

Note that $\ln \left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$

Now, replace $x$ by $-x$ in above equation we get, $-\ln \left(1-x\right)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots$

Hence, adding the two equations we have,
$\ln \left(1+x\right)+\left(-\ln \left(1-x\right)\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots \right)$

Therefore, $\ln \frac{1+x}{1-x}=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots \right)$

Now substitute $x=\frac{m-n}{m+n}$ in above equation to get

$\ln \left\{\frac{1+\left(\frac{m-n}{m+n}\right)}{1-\left(\frac{m-n}{m+n}\right)}\right\}=2\left[\left(\frac{m-n}{m+n}\right)+\frac{1}{3}{\left(\frac{m-n}{m+n}\right)}^3+\frac{1}{5}{\left(\frac{m-n}{m+n}\right)}^5+\cdots \right]$

Simplify the L.H.S.
$\ln \left(\frac{m}{n}\right)=2\left[\left(\frac{m-n}{m+n}\right)+\frac{1}{3}{\left(\frac{m-n}{m+n}\right)}^3+\frac{1}{5}{\left(\frac{m-n}{m+n}\right)}^5+\cdots \right]=$

Hence the correct option is A.