What is the sum of the geometric series 1 + 1/3 + 1/9 + 1/27 ...

The geometric series takes the form of ax^n where n is a number from 0 to infinity with 0 being the first number in the series. 

The sum of this series is S = a + ax + ax^2 + ... 

If we break this series when n is N then the sum of this smaller segment is S_{N}: S_{N} = a + ax + ax^2 + ... + ax^N

If we multiply this by x we have: xS_{N} = ax + ... ax^(N+1)

Then by substracting xS_{N} from S_{N} we have: S_{N}(1 - x) = a(1 - x^{N})

Hence, the sum of this segment is:  S_{N} = (a*(1 - x^{N}))/(1 - x)

Tending N to infinity we get: S = a/(1 - x) In our case a = 1 and x = 1/3 so S = 1/(1 - 1/3) = 1/(2/3) = 3/2

So the answer to the question is 3/2.