Question

How do you find the sum of the infinite geometric series `Sigma 2(-2/3)^n` from n=0 to `oo`?

Answer 1

`6/5`

Explanation:

First, take the `2` out of the sigma so it's easier to deal with:

`2sum(-2/3)^n`

This is in the form:

`sum(u)^n` where `|u|<1`. That means, that `sumu^n=1/(1-u)`

Our `u`, in this case, is `-2/3`. Simply plug into the formula.

`2(1/(1-(-2/3)))=2(1/(5/3))=2(3/5)=6/5`