>>15
Thanks for not bothering with the analytic proof, shithead.
Define a_n as the sequence a_n = sum(9/10^i) for i = 1 to n
Giving (0.9, 0.99, 0.999, ...), which approaches .9999... as n approaches infinity.
A sequence is said to converge to 'a' if for every epsilon > 0, there exists a natural number 'n' such that for all m > n, |a - a_m| < epsilon.
Clearly, |1 - a_n| = 10^-n
If epsilon is greater than or equal to 1, we can choose n = 1, since m > n implies
10^-m = |1 - a_m| < |1 - a_n| = 10^-1 < 1
If epsilon is less than 1, let l = floor(log_10(epsilon) - 1), n = abs(l) and m > n.
n > log_10(epsilon) since epsilon < 1 => log_10(epsilon) < 0
Since log_10(epsilon) < 0 < n < m, it can be said
log_10(epsilon) > -n > -m => 10^(log_10(epsilon)) = epsilon > 10^-n > 10^-m
So for epsilon less than 1, n = abs(floor(log_10(epsilon) - 1)), m > n => |1 - a_m| < epsilon
Since for every epsilon > 0 there exists an 'n' such that m > n => |1 - a_m| < epsilon, we can say that the sequence a_n converges to 1.
QED sluts