Question

How do you find the sum of the infinite geometric series `1/8+1/32+1/128+...`?

Answer 1

`1/6`

Explanation:

First term, `a`, is `1/8`.
Common ratio, `r`, is `1/4`.
Sum of a G.P. to infinity is `a/(1-r)` which is `(1/8)/(1-1/4)=1/6`