- #1

- 6

- 0

^{-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.