Question

If S denotes the sum of infinity and ${S_n}$ the sum of n terms of the series.
$1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+....,$ such that $\text{S}-{{\text{S}}_{\text{n}}}<\dfrac{1}{1000},$ then show that least value of n is 11.

Answer 1

We know that. Sum of infinite G.P.
$\text{S}=\dfrac{a}{1-\text{r}},$ where, a = first term of G.P
r = common ratio of infinite G.P.
and sum of n terms of series,
${{\text{S}}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r},$ where a = first term of G.P
$r\ \text{=}\$ common ration
$n\ =\$ terms of series.
Using given series find the difference of S and ${S_n}$, then using given inequality show $n \ge 11$

Complete step by step solution:
Given series,
$1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+......$
Now,
We know that,
Sum of infinite G.P,$\text{S}=\dfrac{a}{1-r}$--------(A)
Where, $a=$ first term of G.P.
$r\ =$ common ratio of finite G.P
Given that from the series.
$a=1$
$r={\scriptstyle\dfrac{1}{2}}$
Now,
$\text{S}=\dfrac{1}{1-{\scriptstyle\dfrac{-1}{2}}}$
$\Rightarrow \text{S}=\dfrac{1}{\dfrac{2-1}{2}}$
$\Rightarrow \text{S}\ \text{=}\ \dfrac{1}{{\scriptstyle\dfrac{1}{2}}}$
$\Rightarrow \text{S}\ \text{=}\ \text{2}\text{.}$ ---------(1)
Now,
We also know that,
Sum of $n$ terms of series, ${{\text{S}}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$ -----(B)
Where, $a=$ first term of series
$r\ =\$ common rat 20.
$n=$ terms of series.
Given that from the series.
$a=1$
$r\ =\ {\scriptstyle{}^{1}/{}_{2,}}$
From equation (B).
${{\text{S}}_{n}}=\dfrac{1\left( 1-{{\left( {\scriptstyle\dfrac{1}{2}} \right)}^{n}} \right)}{1-{\scriptstyle\dfrac{1}{2}}}$
$\Rightarrow {{\text{S}}_{n}}=\dfrac{1-{{\left( {\scriptstyle\dfrac{1}{2}} \right)}^{n}}}{\dfrac{1}{2}}$
$\Rightarrow {{\text{S}}_{n}}=\left[ 1-{{\left( {\scriptstyle\dfrac{1}{2}} \right)}^{n}} \right]2$
$\Rightarrow {{\text{S}}_{n}}=2-{{\left( \dfrac{1}{2} \right)}^{n}}.2$ $\left( \therefore {{\left( \dfrac{1}{x} \right)}^{n}}={{x}^{-n}} \right)$
$\Rightarrow {{\text{S}}_{n}}=2-{{2}^{-n}}.2$
$\Rightarrow {{\text{S}}_{n}}=2-{{2}^{-n+1}}$
$\Rightarrow {{\text{S}}_{n}}=2-\dfrac{1}{{{2}^{n-1}}}$----------(2)
According to question
From equation (1) and (2).
$\text{S-}{{\text{S}}_{n}}<\dfrac{1}{1000}$
$\Rightarrow 2-\left( 2-\dfrac{1}{{{2}^{n-1}}} \right)<\dfrac{1}{2000}$
$\Rightarrow 2-2+\dfrac{1}{{{2}^{n-1}}}<\dfrac{1}{2000}$
$\Rightarrow \dfrac{1}{{{2}^{2-1}}}<\dfrac{1}{1000}$
-we know that when $\dfrac{1}{x}<\dfrac{1}{y}$ then, $x>y$
So,
$\Rightarrow \dfrac{1}{{{2}^{n-1}}}<\dfrac{1}{1000}$
$\Rightarrow {{2}^{n-1}}\ge 1000$
Now
${{2}^{10}}=32\times 32=1024.$
$\therefore \ n-1\ge 10\ \text{or}\ \text{n}\ge \text{11}\text{.}$

Hence, the least value is 11.

Note:
When the ratio of preceding term and succeeding term is common then the series is called geometric progression (G.P). The sum of infinite Geometric Progression is possible if and only if $|r|<1$ then only sequence will be convergent otherwise it’ll be divergent.