Question

The value of $\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\dots \dots$ is equal to

• A. ${\log }_{e}2$
• B. ${\log }_{e}3$
• C. $lo{g}_e\left(\frac{4}{e}\right)$
• D. ${\log }_e\left(\frac{3}{e}\right)$

Answer 1

The correct option is C $lo{g}_e\left(\frac{4}{e}\right)$

$\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\dots .=\frac{2-1}{1.2}-\frac{3-2}{2.3}+\frac{4-3}{3.4}-\frac{5-4}{4.5}+\dots .$
$=1-\frac{1}{2}-\frac{1}{2}+\frac{1}{3}+\frac{1}{3}-\frac{1}{4}-\frac{1}{4}+\frac{1}{5}+\frac{1}{5}-\dots .=-1+2(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots )$
$=-1+2\log 2=\log 4-\log e=\log \frac{4}{e}$