A) \[{{e}^{2}}\]
B) \[{{\log }_{e}}2+1\]
C) \[{{\log }_{e}}2-1\]
D) \[1-{{\log }_{e}}2\]
Correct Answer: D
Solution :
We have, the given series \[{{\log }_{4}}2-{{\log }_{8}}2+{{\log }_{16}}2-...\infty \] \[=\frac{\log 2}{\log 4}-\frac{\log 2}{\log 8}+\frac{\log 2}{\log 16}-....\infty \] \[=\frac{\log 2}{2\log 2}-\frac{\log 2}{3\log 2}+\frac{\log 2}{4\log 2}-...\infty \] \[=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-...\infty \] \[=1-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-...\infty \] \[1-\left[ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...\infty \right]\] \[=1-{{\log }_{e}}2\]
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