If $S$ denotes the sum to infinite and $S_n$ 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 

Given series $=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$

Here, $a=1,r=\frac{1}{2}$

$\therefore S=\frac{1}{1-\frac{1}{2}}=2\dots \text{ (i) }\left[S_{\mathrm{\infty }}=\frac{a}{1-r}\right]$ and $S_n=\frac{1\left[1-{\left(\frac{1}{2}\right)}^n\right]}{1-\frac{1}{2}}\left[\because S_n=\frac{a\left(1-r^n\right)}{1-r},r<1\right]$

$=2\left(1-\frac{1}{2^n}\right)$   (ii)

It is given that, $S-S_n<\frac{1}{1000}$

$\Rightarrow 2-2\left(1-\frac{1}{2^n}\right)<\frac{1}{1000}$  [from Eqs. (i) and (ii)] 

$\begin{array}{l}\Rightarrow 2-2+\frac{2}{2^n}<\frac{1}{1000}\\ \frac{1}{2^{n-1}}<\frac{1}{1000}\\ \Rightarrow 2^{n-1}>1000\\ \Rightarrow n-1\ge 10\\ \Rightarrow n\ge 11\end{array}\left[\because \frac{1}{a}<\frac{1}{b}\Rightarrow a>b\right]$

So, the least value of $n$ is 11 .