The value of (1)/( _3 e)+(1)/( _3 e^2)+(1)/( _3 e^4)+ . up to...

\frac{1}{\log _3 \mathrm{e}}+\frac{1}{\log _3 \mathrm{e}^2}+\frac{1}{\log _3 \mathrm{e}^4}+\ldots=\frac{1}{\log _3 \mathrm{e}}+\frac{1}{2 \log _3 \mathrm{e}}+\frac{1}{4 \log _3 \mathrm{e}}+\ldots \cdot\left(\right.$Since, \left.\log _{\mathrm{a}} \mathrmb^{\mathrm{m}}=\mathrm{m} \mathrm{log}_{\mathrm{a}} \mathrm{b}\right)=\log _{\mathrm{e}} 3+\frac{\log _{\mathrm{e}} 3}{2}+\frac{\log _{\mathrm{e}} 3}{4}+\ldots . \quad\left(\right.$ Since, \left.\log _{\mathrm{a}} \mathrm{b}=\frac{1}{\log _{\mathrm{b}} \mathrm{a}}\right)=\log _{\mathrm{e}} 3\left(1+\frac{1}{2}+\frac{1}{4}+\ldots.\right)=\log _{\mathrm{e}} 3\left(\frac{1}{1-\frac{1}{2}}\right)\left(\because 1, \frac{1}{2}, \frac{1}{4} \ldots\right.$is G.P. with \left.\mathrm{a}=1, \mathrm{r}=\frac{1}{2}\right)=\log _{\mathrm{e}} 3(2)=2 \log _{\mathrm{e}} 3=\log _{\mathrm{e}} 3^2=\log _{\mathrm{e}} 9$.