Originally Posted by a-j

Here is another way
.9r = .999999999... = 9*(.1111111111...)
.9r = 9*sum(i=1 to inf, (.1)^i) <= read the sum of i=1 to infinity of (.1)^i

sum(i=1 to inf, (.1)^i) = an infinite geometric sum (.1^0 + .1^1 + .1^2 + .1^3... ) - 1
The sum (.1^0 + .1^1 + .1^2 + .1^3... ) is well known to converge to 1/(1-r) where r = .1, so,
.9r = 9*(1/(1-.1) - 1)
.9r = 9*(1/.9 - 1)
.9r = 9*(1/(9/10) - 1)
.9r = 9*(10/9 - 9/9)
.9r = 9*(1/9)
.9r = 1