Question

How do you find the sum of the infinite geometric series `Sigma (0.4)^n` from n=0 to `oo`?

Answer 1

`5/3`

Explanation:

Where `absr<1`, note that `sum_(n=0)^oor^n=1/(1-r)`.

Here, since `abs0.4<1`, we see that

`sum_(n=0)^oo(0.4)^n=1/(1-0.4)=1/0.6=1/(3//5)=5/3`