# For the geometric series 4+1/4+(1/4)^2+(1/4)^3+.... find the least value of n such that |Sn-Sinfinity|<0.0001

### 1 Answer | Add Yours

The series should be

`S_n = 1 + 1/4 + 1/4^2+1/4^3+........+1/4^(n-1)`

with common ratio of `r = 1/4` and a = 1

then for a geometric series,the sum of first n terms is,

`S_n = sum_(k=0)^(n-1)ar^k = a(1-r^n)/(1-r)`

the sum if n goes to inifinity is,

`S_oo = sum_(k=0)^(n-1)ar^k = a/(1-r)`

for our case,

`S_n = sum_(k=0)^(n-1)(1/4)^k = (1-(1/4)^n)/(1-(1/4))`

`S_oo = sum_(k=0)^(n-1)(1/4)^k = 1/(1-(1/4))`

therefore `|s_n - s_oo| = |4/3 - 4/3(1-(1/4)^n)|`

`|s_n - s_oo| = 4/3|1-(1-(1/4)^n)|`

`|s_n - s_oo|= 4/3(1/4)^n`

Our requirement is,

`|s_n - s_oo| < 0.0001`

`4/3(1/4)^n<0.0001`

`(1/4)^n<0.000075`

taking the `log_10` of each side

`n log_10(1/4) < log_10(0.000075)`

`n * (-0.6021) lt -4.1249`

`n gt 6.85` (Dividing by negative changes the inequality)

Therefore n should be greater than 6.85. Since n is an integer, n should be greater than 7.

The minimum value of n is 7.

**Sources:**