Question

How do you find the sum of the infinite geometric series `Sigma (1/10)^n` from n=0 to `oo`?

Answer 1

`sum_(n=0)^oo (1/10)^n = 10/9`

Explanation:

The sum of a geometric series is given by:

`sum_(n=0)^oo x^n = 1/(1-x)` for `abs x < 1`

In our case:

`x = 1/10 < 1`

so:

`sum_(n=0)^oo (1/10)^n = 1/(1-1/10) = 10/9`