[evanb]

Sequences can approach things. The sequence 0.9, 0.99, 0.999, 0.9999 and so on asymptotically approaches 1. The difference between 1 and the Nth term in the sequence is 1e-N, which goes to 0 with N.

0.999...[forever] is not a sequence, it is a number. Numbers have values, they don't approach things. The misleading part is that 'forever' is not something about evolution or the passage of time. It's not 'happening' or 'sequential' like the sequence. There is no 'and then another 9'. All the 9s are really there, at once. And it is closer to 1 than any term in the sequence. Since the sequence gets closer and closer to 1, converging to it asymptotically, 0.999...[forever] cannot differ from 1; if it did the sequence wouldn't converge.

[eszed]

Thank you, and everyone else who answered (I hope they see this reply). Your distinction between "sequence" and "number", along with the mathematics of 0.333... = 1/3, convinced me - and my mind is successfully blown.

Follow-up: is it the same for other repeating sequence-looking numbers? As in, would 0.9333... = 0.94?

[eigenket]

Its true of numbers whose with a decimal representation which ends in an infinite string of 9s, so 0.939999... = 0.94. This is because we write numbers in base 10. If you write numbers in base 2 its equal to numbers whose binary representation ends with an infinte number of 1s e.g. 0.11111... (base 2) = 1.

[evanb]

0.9333... is equal to 9/10 + 1/30. To get 9/10 + 4/100 I think what you're aiming at is 0.93999...