Question

If $y>0,$ then prove that
${\log }_{e}y=2[\left(\frac{y-1}{y+1}\right)+\frac{1}{3}{\left(\frac{y-1}{y+1}\right)}^3+\frac{1}{5}{\left(\frac{y-1}{y+1}\right)}^5+...]$ and calculate ${\log }_{e}2$ to three places of decimal.

Answer 1

We know $x=\frac{y-1}{y+1},$ we get ${\log }_e\left(\frac{1+\frac{y-1}{y+1}}{1-\frac{y-1}{y+1}}\right)=2[\frac{y-1}{y+1}+\frac{1}{3}{\left(\frac{y-1}{y+1}\right)}^2+\frac{1}{5}{\left(\frac{y-1}{y+1}\right)}^5+...]$
${\log }_{e}y=2[\frac{y-1}{y+1}+\frac{1}{3}{\left(\frac{y-1}{y+1}\right)}^2+\frac{1}{5}{\left(\frac{y-1}{y+1}\right)}^5+...]$
putting $y=2,$ we get ${\log }_{e}2=2[\frac{1}{3}+\frac{1}{3}{\left(\frac{1}{3}\right)}^3+\frac{1}{5}{\left(\frac{1}{5}\right)}^5+...]=\frac{2}{3}[1+{\left(\frac{1}{3}\right)}^3+\frac{1}{5}{\left(\frac{1}{3}\right)}^4+...]=0.693$