How do you find the sum of the infinite geometric series 3 + 1 + 1/3 + 1/9 + ...?

1 Answer
Jan 29, 2016

4.5

Explanation:

Note that a/(1-a) yields the infinite sum a+a^2+a^3+a^4+ ....
(One can do the specified division to see why this is true.) Substituting a=1/3, one gets that the infinite sum 1/3+1/3^2+1/3^3+ ... is equal to (1/3/(1-1/3)) = (1/3)/(2/3) = 1/2. So, adding one to this infinite sum yields 3/2. Note that the given infinite sum is three times our sum, i.e. one can factor 3 out of each term, to get 3(1+1/3+1/3^2+1/3^3+ ...), so the sum of the given series is 3(3/2) =9/2 = 4.5.