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Question: Answered & Verified by Expert
$\frac{1}{2 \cdot 3}+\frac{1}{4 \cdot 5}+\frac{1}{6 \cdot 7}+\frac{1}{8 \cdot 9}+\ldots$ is equal to
MathematicsBasic of MathematicsAP EAMCETAP EAMCET 2013
Options:
  • A $\log \left(\frac{2}{e}\right)$
  • B $\log \left(\frac{e}{2}\right)$
  • C $\log (2 e)$
  • D $e-1$
Solution:
1867 Upvotes Verified Answer
The correct answer is: $\log \left(\frac{e}{2}\right)$
$\begin{aligned} & \frac{1}{2 \cdot 3}+\frac{1}{4 \cdot 5}+\frac{1}{6 \cdot 7}+\frac{1}{8 \cdot 9}+\ldots \infty \\ & =\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)+\ldots \infty \\ & =1-\log _e 2 \\ & =\log _e e-\log _e 2 \\ & =\log \left(\frac{e}{2}\right)\end{aligned}$

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