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The value of $\frac{1}{\log _{3} e}+\frac{1}{\log _{3} e^{2}}+\frac{1}{\log _{3} e^{4}}+\ldots$ up to infinite
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The correct answer is:
$\log _{e} 9$
$\frac{1}{\log _{3} \mathrm{e}}+\frac{1}{\log _{3} \mathrm{e}^{2}}+\frac{1}{\log _{3} \mathrm{e}^{4}}+\ldots .$
$=\frac{1}{\log _{3} \mathrm{e}}+\frac{1}{2 \log _{3} \mathrm{e}}+\frac{1}{4 \log _{3} \mathrm{e}}+\ldots .\left(\right.$ Since, $\left.\log _{\mathrm{a}} \mathrm{b}^{\mathrm{m}}=\mathrm{m} \log \mathrm{a} \mathrm{b}\right)$
$=\log _{\mathrm{e}} 3+\frac{\log _{\mathrm{e}} 3}{2}+\frac{\log _{\mathrm{e}} 3}{4}+\ldots \cdot\left(\right.$ Since $\left.\log _{\mathrm{a}} \mathrm{b}=\frac{1}{\log _{\mathrm{b}} \mathrm{a}}\right)$
$\quad=\log _{\mathrm{e}} 3\left(1+\frac{1}{2}+\frac{1}{4}+\ldots .\right)$
$=\log _{\mathrm{e}} 3\left(\frac{1}{1-\frac{1}{2}}\right)\left(\because 1, \frac{1}{2}, \frac{1}{4} \ldots .\right.$ is G.P. with $\left.\mathrm{a}=1, \mathrm{r}=\frac{1}{2}\right)$
$=\log _{\mathrm{e}} 3(2)=2 \log _{\mathrm{e}} 3=\log _{\mathrm{e}} 3^{2}=\log _{\mathrm{e}} 9$
$=\frac{1}{\log _{3} \mathrm{e}}+\frac{1}{2 \log _{3} \mathrm{e}}+\frac{1}{4 \log _{3} \mathrm{e}}+\ldots .\left(\right.$ Since, $\left.\log _{\mathrm{a}} \mathrm{b}^{\mathrm{m}}=\mathrm{m} \log \mathrm{a} \mathrm{b}\right)$
$=\log _{\mathrm{e}} 3+\frac{\log _{\mathrm{e}} 3}{2}+\frac{\log _{\mathrm{e}} 3}{4}+\ldots \cdot\left(\right.$ Since $\left.\log _{\mathrm{a}} \mathrm{b}=\frac{1}{\log _{\mathrm{b}} \mathrm{a}}\right)$
$\quad=\log _{\mathrm{e}} 3\left(1+\frac{1}{2}+\frac{1}{4}+\ldots .\right)$
$=\log _{\mathrm{e}} 3\left(\frac{1}{1-\frac{1}{2}}\right)\left(\because 1, \frac{1}{2}, \frac{1}{4} \ldots .\right.$ is G.P. with $\left.\mathrm{a}=1, \mathrm{r}=\frac{1}{2}\right)$
$=\log _{\mathrm{e}} 3(2)=2 \log _{\mathrm{e}} 3=\log _{\mathrm{e}} 3^{2}=\log _{\mathrm{e}} 9$
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