# Thread: What's wrong with this proof...

1. Let me teach you some of the basics of a geometric series...

The difference between a sequence and a series is:

Series: $\displaystyle S_n = 1 + 2 + 4 + 8 + 16 + ...$
Sequence: $\displaystyle S_n = 1 ; 2 ; 4 ; 8 ; 16 ; ...$

A geometric series is formed by multiplying each term by a constant.

For example: $\displaystyle S_n = 1 + 2 + 4 + 8 + 16 + ...$
Could also be written as: $\displaystyle S_n = (1)(2^0) + (1)(2^1) + (1)(2^2) + (1)(2^3) + (1)(2^4) + ...$

The n-th term is equal to: $\displaystyle ar^{n - 1}$

$\displaystyle a$ is the starting term, the first term (a = 1 in the example).
$\displaystyle r$ is the common ratio (r = 2 in the example).

You can determine $\displaystyle r$ with the following formula: $\displaystyle r = \frac{T_2}{T_1}$

In series and sequences, the Sigma-sign is very important.

$\displaystyle \sum^{3}_{n = 0} (n + 1) = (0 + 1) + (1 + 1) + (2 + 1) + (3 + 1) = 10$

Now i do hope that's all...

2. There is nothing wrong with the proof, it is in fact true that $\displaystyle .9999.... = 1$.

Note: If this thread starts getting "philosophical" , just be warned, you know I hate that stuff and I will close it.

3. The basic fact you need to recall is: $\displaystyle \sum\limits_{k = J}^\infty {ar^k } = \frac{{ar^J }}{{1 - r}}\quad \text{if}\quad \left| r \right| < 1$.

Example: $\displaystyle \sum\limits_{k = 5}^\infty {3\left( {\frac{4}{5}} \right)^k = \frac{{3\left( {4/5} \right)^5 }}{{1 - \left( {4/5} \right)}}}$

Here is the way a so-called repeating decimal works.
$\displaystyle .9\overline 9 = .9 + .09 + .009 + .0009 + \cdots$
$\displaystyle .9\overline 9 = \frac{9}{{10}} + \frac{9}{{100}} + \frac{9}{{1000}} + \frac{9}{{10000}} + \cdots$
$\displaystyle .9\overline 9 = \frac{9}{{10^1 }} + \frac{9}{{10^2 }} + \frac{9}{{10^3 }} + \frac{9}{{10^4 }} + \cdots$
$\displaystyle .9\overline 9 = \sum\limits_{k = 1}^\infty {9\left( {\frac{1}{{10}}} \right)^k } = \frac{{\left( {9/10} \right)}}{{1 - \left( {1/10} \right)}} = 1$

4. Hello, jtl!

There's nothing wrong with that proof.

Let $\displaystyle N \;=\;0.99999\cdots \;=\;\frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots$

This is a geometric series with first term $\displaystyle a = \frac{9}{10}$ and common ratio $\displaystyle r = \frac{1}{10}$
. . Its sum is: .$\displaystyle N \;=\;\frac{\frac{9}{10}}{1-\frac{1}{10}} \;=\;\frac{\frac{9}{10}}{\frac{9}{10}} \;=\;1$

Therefore: .$\displaystyle 0.99999\cdots \;=\;1$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Another proof

We have: .$\displaystyle N \;=\;0.99999\cdots$

Multiply by 10: .$\displaystyle 10N \;=\;9.99999\cdots$
. . Subtract $\displaystyle N$: . . $\displaystyle N \;=\;0.99999\cdots$

And we have: .$\displaystyle 9N \:=\:9\quad\Rightarrow\quad N \:=\:1$

5. Originally Posted by ThePerfectHacker
There is nothing wrong with the proof, it is in fact true that $\displaystyle .9999.... = 1$.

Note: If this thread starts getting "philosophical" , just be warned, you know I hate that stuff and I will close it.
Don't worry, with my level of math, I don't think I'll last in a philosophical battle.

6. Thanks everyone, Soraoban and Plato and others, it's clear now.

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