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 the least value of n is-

• A. $8$
• B. $9$
• C. $10$
• D. $11$

Answer 1

The correct option is D $11$

By using Sum of GP $=\frac{a(1-r^n)}{1-r}$ when $r<1$ we have

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

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

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

Now $2^9=512$

$\therefore n-1>9$ or $n>10$

Hence the least value is $11$.