Question

IF Sn=1+1/2+1/2^2+.......+1/(2^n-1),find the least value of n for which 2-Sn<1/100?

Answer 1

`n=8`

Explanation:

`S_n=1+1/2+1/2^2+1/2^3+…+1/2^(n-1)`

The above sum is a geometric series `S_n=a+ar+ar^2+ar^3+…+ar^(n-1)` with `a=1` and `r=1/2`.

The sum of a geometric series is given by `S_n=(a(1-r^n))/(1-r)`.

Substitute the values in for the geometric series given in the problem to get `S_n=1+1/2+1/2^2+1/2^3+…+1/2^(n-1)=(1-(1/2)^n)/(1-1/2)`.

This simplifies down to `2-2(1/2)^n=2-2*2^-n=2-2^(1-n)`.

We need to find the smallest integer `n` such that `2-S_n=2-(2-2^(1-n))<1/100`.

Manipulate the equation above to obtain `2^(1-n)<1/100`. Now, just substitute values to get that all integers `n>7` will work. Thus, the smallest integer `n` is `8`.