Question

Find the sum of the series
$\frac{1}{2\cdot 3}+\frac{1}{4\cdot 5}+\frac{1}{6\cdot 7}+...$

• A. $log\frac{e}{2}$
• B. $log\frac{e}{4}$
• C. $log\frac{2}{3}$
• D. $log\frac{2}{4}$

Answer 1

The correct option is A $log\frac{e}{2}$

$\frac{1}{2\cdot 3}+\frac{1}{4\cdot 5}+\frac{1}{6\cdot 7}$

$\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}+...$

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

$=1-(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5})$

$=1-log2=loge-log2$

$=log\frac{e}{2}$