PROOF that 0.99999... = 1

From: mike3 (mike4ty4_at_yahoo.com)
Date: 12/01/04 

Date: 1 Dec 2004 13:50:07 -0800

Hi.

Theorem: 0.9999... = 1

Proof:

A number x = 0.aaaaaa..., where 0 <= a <= 9 is

x = sum_{n=1...inf} a/(10^n)

If a = 9, then

x = sum_{n=1...inf} 9/(10^n)

Then, we have the sequence of partial sums

S_n = sum_{i=1...n} 9/(10^n).

We can then derive the _general formula_ as follows:

S_1 = 9/10
S_2 = 9/10 + 9/100 = 99/100
S_3 = 9/10 + 9/100 + 9/1000 = 999/1000
...
S_n = ((10^n)-1)/(10^n)
...

Then we take the limit as n -> inf:

lim_{n->inf} ((10^n)-1)/(10^n).

Direct susbtitution produces the indeterminate form inf/inf, so we rewrite
the limit:

lim_{n->inf} ((10^n)-1)/(10^n) = lim_{n->inf} 1 - (1/(10^n))
                               = lim_{n->inf} 1 - lim_{n->inf} (1/(10^n))
                               = 1 - 0
                               = 1

Thus, 0.9999... = 1. QED.

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