Question

The value of $\frac{1}{{\log }_{3}e}+\frac{1}{{\log }_3{e}^2}+\frac{1}{{\log }_3{e}^4}+......$ up to infinite terms

Answer 1

${\log }_{e}3+\frac{1}{2}{\log }_{e}3+\frac{1}{4}{\log }_{e}3+..........$

${\log }_{e}3(1+\frac{1}{2}+\frac{1}{4}+........)$

${\log }_{e}3\times \frac{1}{1-\frac{1}{2}}$

${\log }_{e}3\times \frac{1}{\frac{1}{2}}$

$2{\log }_{e}3={\log }_{e}9$