Question

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

Answer 1

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`.