Question

The sum of the series ${\log }_{4}2-{\log }_{8}2+{\log }_{16}2-...$ is

• A. $e^2$
• B. ${\log }_{e}2+1$
• C. ${\log }_{e}3-2$
• D. $1-{\log }_{e}2$

Answer 1

The correct option is D $1-{\log }_{e}2$

The given series can be written as$\frac{\ln \left(2\right)}{2\ln \left(2\right)}-\frac{\ln \left(2\right)}{3\ln \left(2\right)}+\frac{\ln \left(2\right)}{4\ln \left(2\right)}...\infty$

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

$=-[-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}...\infty ]$

$=-\left[\ln \right(1+1)-1]$

$=1-\ln \left(2\right)$

$=1-{\log }_e\left(2\right)$

Hence, option 'D' is correct.