Question

How do you use a geometric series to prove that `0.999…=1`?

Answer 1

We can write
`0.999...=9/10+9/100+9/1000+cdots`
`=9/10+9/10(1/10)+9/10(1/10)^2+cdots`
`=sum_{n=0}^infty 9/10(1/10)^n`,
which is a geometric series with
`a=9/10` and `r=1/10`.

So, the sum is
`a/{1-r}={9/10}/{1-1/10}=1`