Question 391682
The number you're referring to is 1.99999....there are probably a dozen ways to prove this is equal to 2.


We could consider the infinite series {{{sum(9*10^(-k), k = 1, infinity) = 9/10 + 9/100 + 9/1000}}} + ...If you know that the sum of the terms of a convergent geometric sequence {{{ar^i}}} is {{{a/(r-1)}}} we can replace a = 9, and r = 10 to get 9/9 or 1. Finally, add 1 (since the number is 1.9999 instead of .9999) to get 2.


Another possible way works like this:


1/9 = .11111....
9/9 = 9(.11111....) = .99999....


However 9/9 = 1 so .99999.... = 1, and 1.99999.... = 2 = 2/1.