Question

For the infinite series $1-\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}-\frac{1}{32}+\frac{1}{54}-\frac{1}{128}-....$ let $S$ be the (limiting) sum. Then $S$ equals

• A. $0$
• B. $\frac{2}{7}$
• C. $\frac{6}{7}$
• D. $\frac{9}{32}$
• E. $\frac{27}{32}$

Answer 1

The correct option is B $\frac{2}{7}$

Combine the terms in threes, to get the geometric series
$\frac{1}{4}+\frac{1}{32}+\frac{1}{256}+....;S=\frac{\frac{1}{4}}{1-\frac{1}{8}}=\frac{2}{7}$ or
rearrange the terms into three series:
$1+\frac{1}{8}+\frac{1}{64}+...-\frac{1}{2}-\frac{1}{16}-\frac{1}{128}-....,-\frac{1}{4}-\frac{1}{32}-\frac{1}{256}-....$
$S_1=\frac{1}{1-\frac{1}{8}}=\frac{8}{7};S_2=\frac{-\frac{1}{2}}{1-\frac{1}{8}}=-\frac{4}{7};S_3=\frac{-\frac{1}{4}}{1-\frac{1}{8}}=-\frac{2}{7};\therefore S=\frac{2}{7}$