Represent .999... by (1-10-n), limit as n>infinity. Then (1-10-n)10n, limit as n>infinity = 1/e (binomial expansion) (1+10-n)10n, limit as n>infinity = e (1)10n, limit as n>infinity=1 I think .999... is a transcendental number that can in almost every case be treated as equal to 1. Any solution to a finite algebraic equation that approximates 1 but is less than 1 will always be less than .999... If .999... is transcendental then so is every other recurring decimal because they are arbitrarily close to a fraction but not equal to that fraction. The above equation seems to converge with the number of digits of accuracy approximately equal to n.