Question

What is the sum of the infinite geometric series 9+6+4+...?

Answer 1

`S=27`

Explanation:

The series can be written as
`S=9+6+4+8/3+16/9+32/27+...+4 (2/3)^n`
`S=15+4sum_{i=0}^{i=infty}(2/3)^i`
Now we will use a polynomial identity which says
`(1-x^{n+1})/(1-x) = 1+x+x^2+x^3+...+x^{n}`
So for `n`
`sum_{i=0}^{i=n}(2/3)^i = (1-(2/3)^{n+1})/(1-(2/3))`
but as `n->infty`, `(2/3)->0`, because `(2/3)<1`
Finally we get
`sum_{i=0}^{i=infty}(2/3)^i = 1/(1-(2/3)) = 3`
Finally we obtain
`S = 15+4times3 = 27`