Question

If $S$ denotes the sum to infinity and $S_n$ denotes the sum of $n$ terms of the series $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots ,$ such that $S-S_n<\frac{1}{1000},$ then the least value of $n$ is

• A. 8
• B. 9
• C. 10
• D. 11

Answer 1

Answer: (D)

We have, $S=\frac{1}{1-\frac{1}{2}}=2$ and
$S_n=\frac{\left(1-1/2^n\right)}{(1-1/2)}=2\left(1-\frac{1}{2^n}\right)=2-\frac{1}{2^{n-1}}$
$\therefore S-S_n<\frac{1}{1000}\Rightarrow \frac{1}{2^{n-1}}<\frac{1}{1000}$
$\Rightarrow 2^{n-1}>1000$
$\Rightarrow n-1\ge 10$
$\Rightarrow n\ge 11$
Hence, the least value of $n$ is 11