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$\frac{1}{2}-\frac{1}{2 \cdot 2^2}+\frac{1}{3 \cdot 2^3}-\frac{1}{4 \cdot 2^4}+\ldots$ is equal to
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Verified Answer
The correct answer is:
$\log _e\left(\frac{3}{2}\right)$
Given,
$\frac{1}{2}-\frac{1}{2 \cdot 2^2}+\frac{1}{3 \cdot 2^3}-\frac{1}{4 \cdot 2^4}+\ldots$
On Comparing with
$\log _e(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots \infty$
Put $x=\frac{1}{2}$ on both sides, we get
$\begin{aligned} \frac{1}{2}-\frac{1}{2 \cdot 2^2}+\frac{1}{3 \cdot 2^3}-\frac{1}{4 \cdot 2^4}+\ldots & =\log _e\left(1+\frac{1}{2}\right) \\ & =\log _e\left(\frac{3}{2}\right)\end{aligned}$
$\frac{1}{2}-\frac{1}{2 \cdot 2^2}+\frac{1}{3 \cdot 2^3}-\frac{1}{4 \cdot 2^4}+\ldots$
On Comparing with
$\log _e(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots \infty$
Put $x=\frac{1}{2}$ on both sides, we get
$\begin{aligned} \frac{1}{2}-\frac{1}{2 \cdot 2^2}+\frac{1}{3 \cdot 2^3}-\frac{1}{4 \cdot 2^4}+\ldots & =\log _e\left(1+\frac{1}{2}\right) \\ & =\log _e\left(\frac{3}{2}\right)\end{aligned}$
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