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
$S_{n}=\frac{\left(1-1 / 2^{n}\right)}{(1-1 / 2)}=2\left(1-\frac{1}{2^{n}}\right)=2-\frac{1}{2^{n-1}}$
$\therefore S-S_{n}<\frac{1}{1000} \Rightarrow \frac{1}{2^{n-1}}<\frac{1}{1000}$
$\Rightarrow \quad 2^{n-1}>1000$
$\Rightarrow n-1 \geq 10$
$\Rightarrow n \geq 11$
Hence, the least value of $n$ is 11