- #106
Caramon
- 133
- 5
Char. Limit said:Here's a proof that .999...=1.
.999... can be written as the infinite sum as follows:
[tex].9 + .09 + .009 + .0009 + .00009 + ... = .9 \sum_{n=0}^\infty \left(\frac{1}{10}\right)^n[/tex]
Now, evaluating the sum on the right, we use the fact (proven below) that...
[tex]\sum_{n=0}^\infty r^n = \frac{1}{1-r}[/tex]
for all r with a magnitude less than 1. Using this fact, we find that...
[tex].9 \sum_{n=0}^\infty \left(\frac{1}{10}\right)^n = .9 \frac{1}{1-.1} = \frac{.9}{.9}=1[/tex]
Now, to prove that fact that we used, note the proof below:
[tex]S = \sum_{k=0}^{n-1} a r^k = a + a r + a r^2 + a r^3 + a r^4 + ... + a r^{n-1}[/tex]
[tex]rS = a r + a r^2 + a r^3 + a r^4 + a r^5 + ... + a r^n[/tex]
[tex]S - rS = a - a r^n = a (1 - r^n)[/tex]
[tex]S(1-r) = a (1 - r^n)[/tex]
[tex]S = \frac{a (1-r^n)}{1-r}[/tex]
Now let n go to infinity. For r with a magnitude less than 1, r^n tends to 0 as n tends to infinity. Thus...
[tex]\lim_{n \rightarrow \infty} S = \lim_{n \rightarrow \infty} \frac{a (1-r^n)}{1-r} = \frac{a}{1-r}[/tex]
Q E D
/thread
This thread is INFINITELY hilarious because people don't need to say anything other than, "oh wow this proof is great."