Question

How do you find the sum of the infinite geometric series `Sigma -3(0.9)^n` from n=0 to `oo`?

Answer 1

`S = sum_(n=0)^oo -3(0.9)^n = -30`

Explanation:

The sum:

`S = sum_(n=0)^oo -3(0.9)^n`

represents a GP with `a=-3` an `r=0.9` so as `|r|<1`we can use the standard GP summation result:

`s_oo = a/(1-r)`

Which gives us:

`S = (-3)/(1-0.9)`
`\ \ = (-3)/(0.1)`
`\ \ = -30`