Question

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

Answer 1

`3sum_{i=0}^{infty}(-1/3)^i = 9/4`

Explanation:

The sum of sequence `{3, -1,1/3,-1/9,1/27,...} = 3{1,-1/3,1/9,-1/27,...}` or more compactly
`S=3sum_{i=0}^{infty}(-1/3)^i = 3(1/(1+1/3)) = 9/4`
We use here the polynomial identity
`(1-x^{n+1})/(1-x) = 1+x+x^2+x^3+...+x^n`. Here `x = -1/3` and `abs x < 1` so `lim_{n->infty} = 1/(1-x)`