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

An infinite geometric series is of the form a*(1 + b + b^2 + b^3 ...)

To find the sum of all the terms in this series we consider the sum, S, where:

S= 1 + b + b^2 + b^3 ... + b^n

where n is some finite integer.

Then, if we multiply all the terms in this sum by a factor b, we get:

S*b= (b + b^2 + b^3 + b^4 ... + b^(n+1)

Subtracting the previous sum from this new sum we get:

S*b - S= b^(n+1) - 1

Factoring out the S from the left hand side of the equality we get:

S(b - 1)= b^(n+1) - 1

Hence, an equation for the sum of the terms in the finite geometric series is:

S= (b^(n+1) - 1)/(b - 1)

But we wanted to find the sum of a finite series, so tending the n to infinity and assuming that the factor b is less than one gives the sum of an infinite geometric series:

sum of an infinite geometric series= a/(1 - b)

where a was the prefactor in the form.

In the original question a was 1 and b was 1/3 so the sum of our infinite geometric series is 1/(1 - 1/3)= 3/2.