Question

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

Answer 1

$S=\frac{1}{1-\left(1/2\right)}=2,$

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

$S-S_n=\frac{1}{2^{n-1}}<\frac{1}{1000}$ or $2^{n-1}\ge 1000$

Now $2^{10}=32\times 32=1024$

$\therefore n-1\ge 10$ or $n\ge 11$

Hence the least value is $11$.