Question

The least value of n ( a natural number), for which the sum S of the series $1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots$ differs from $S_n$ by a quantity $<{10}^{-6}$ is

• A. 21
• B. 20
• C. 19
• D. None

Answer 1

The correct option is A 21

$S=\frac{1}{1-\frac{1}{2}}=2S_n=\frac{1(1-\frac{1}{2^n})}{1-\frac{1}{2}}=2-\frac{1}{2^{n-1}}$
Given $S-S_n<{10}^{-6}\therefore \frac{1}{2^{n-1}}<{10}^{-6}\Rightarrow 2^{n-1}>{10}^6\therefore n-1>6lo{g}_{2}10=\frac{6}{0.3010}\Rightarrow n>20[\because \frac{6}{.3018}<20]\therefore n=21$