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The sum of the series

$\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-........upto\: \infty$

is equal to
Option: 1
$\log_{e}2-1$

Option: 2
$\log_{e}2$

Option: 3
$\log_{e}\left ( 4/e \right )$

Option: 4
$2\log_{e}2$

Answers (1)

best_answer

Using @6 and @7

$\frac{1}{1.2}- \frac{1}{2.3}+ \frac{1}{3.4}\cdot \cdot \cdot \infty$

$\therefore \frac{1}{1.2}= \frac{1}{1}-\frac{1}{2}$

$\frac{1}{2.3}= \frac{1}{2}-\frac{1}{3}$

So that $\left ( 1-\frac{1}{2} \right )-\left ( \frac{1}{2} -\frac{1}{3}\right )+\left ( \frac{1}{3} -\frac{1}{4}\right )-\left ( \frac{1}{4}-\frac{1}{5} \right )\cdot \cdot \cdot \infty$

$= 1-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}+\frac{2}{5}\cdot \cdot \cdot \infty$

$= 1+2\left [ -\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\cdot \cdot \cdot \right ]$

$= 1+2\left [ log2-1 \right ]$ $\left [ \because log\left ( 1-x \right )= x-\frac{x^{2}}{2}+\frac{x^{3}}{3}\cdot \cdot \cdot \right ]$

$= 1=2log2-2$ Put x = 1

$= log{_{e}}^{4}-1$ $log2= 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\cdot \cdot \cdot$

= $log{_{e}}^{4}-log{_{e}}^{e}$ $log2-1= -\frac{1}{2}+\frac{1}{3}\cdot \cdot \cdot \cdot$

$= log_{e}^{\frac{4}{e}}$

Posted by

Suraj Bhandari

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