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$1+\frac{1}{3 \cdot 2^2}+\frac{1}{5 \cdot 2^4}+\frac{1}{7 \cdot 2^6}+\ldots$ is equal to
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The correct answer is:
$\log _e 3$
Given series is $1+\frac{1}{3 \cdot 2^2}+\frac{1}{5 \cdot 2^4}+\frac{1}{7 \cdot 2^6}+\ldots$
$\begin{aligned} & =2\left[\frac{1 / 2}{1}+\frac{(1 / 2)^3}{3}+\frac{(1 / 2)^5}{5}+\ldots\right] \\ & =\log _e\left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right)=\log _e\left(\frac{3 / 2}{1 / 2}\right) \\ & {\left[\because \log _e\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{3}+\ldots\right)\right]} \\ & =\log _e 3 \\ & \end{aligned}$
$\begin{aligned} & =2\left[\frac{1 / 2}{1}+\frac{(1 / 2)^3}{3}+\frac{(1 / 2)^5}{5}+\ldots\right] \\ & =\log _e\left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right)=\log _e\left(\frac{3 / 2}{1 / 2}\right) \\ & {\left[\because \log _e\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{3}+\ldots\right)\right]} \\ & =\log _e 3 \\ & \end{aligned}$
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