The sum of the series (\log \_{4} 2-\log \_{8} 2+\log \_{16}... | Mathematics | SolveeIt

Question

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

$e^2$

$\log _{e} 2$

$\log _{e} 3-2$

$1 - \log_e 2$

Answer

$1 - \log_e 2$

Explanation

Solution

$\log _{4} 2-\log _{8} 2+\log _{16} 2-\ldots$ $=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\ldots\right)-1+1$ $=-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots\right)+1$ $=1-\log (1+1)=1-\log _{e} 2$

Mathematics Question on Sum of First n Terms of an AP

Solution