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 $\mathrm{S}-{\mathrm{S}}_{\mathrm{n}}<\frac{1}{1000}$then the least value of n is 

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

Here, $\mathrm{a}=1,\mathrm{r}=\frac{1}{2}$

$\therefore \mathrm{S}=\frac{1}{1-\frac{1}{2}}=2$ $\left[{\mathrm{S}}_{\mathrm{\infty }}=\frac{\mathrm{a}}{1-\mathrm{r}}\right]$

and ${\mathrm{S}}_{\mathrm{n}}=\frac{1\left[1-{\left(\frac{1}{2}\right)}^{\mathrm{n}}\right]}{1-\frac{1}{2}}\left[\because {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left(1-{\mathrm{r}}^{\mathrm{n}}\right)}{1-\mathrm{r}},\mathrm{r}<1\right]$

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

It is  given  that, $\mathrm{S}-{\mathrm{S}}_{\mathrm{n}}<\frac{1}{1000}$

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

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2-2+\frac{2}{2^{\mathrm{n}}}<\frac{1}{1000}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{1}{2^{\mathrm{n}-1}}<\frac{1}{1000}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{2}^{\mathrm{n}-1}>1000\left[\because \frac{1}{\mathrm{a}}<\frac{1}{\mathrm{b}}\Rightarrow \mathrm{a}>\mathrm{b}\right]\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{n}-1\ge 10\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{n}\ge 11\end{array}$

So, the  least  value  of n is 11.