Question on Geometric series

[Johnny5]

Is the following derivation correct?

 

We want so compute the sum of the following infinite series:

 

[math] 1+x+x^2+x^3+x^4+....x^n+... [/math]

 

If x is greater than one, the series clearly diverges, but the interval where x is greater than or equal to zero is not so clear.

 

Assume that the series is convergent, and the sum is S.

 

Therefore:

 

[math] S = 1+x+x^2+x^3+x^4+....x^n+... [/math]

 

Therefore:

 

[math] xS = x+x^2+x^3+x^4+....x^n+... [/math]

 

Therefore:

 

[math] S-Sx=1 [/math]

 

By the distributive axiom of algebra:

 

[math] S-Sx=S(1-x) [/math]

 

By the transitive property of equality, it now follows that:

 

[math] 1=S(1-x) [/math]

 

 

Now, as long as

 

[math] (1-x) \neq 0 [/math],

 

we can divide both sides of the equation above by (1-x).

 

Assume that not [(1-x)=0].

 

So...

 

[math] S = \frac{1}{1-x} [/math]

 

The formula on the right, is the famous formula for the sum of the terms of a geometric series, as seen here:

 

Geometric Series

 

 

So my first question, is whether or not the derivation is correct.

 

Thank you

[Dave]

Looks fine to me. That's certainly we way I derive it.

 

(It works for [math]|x| < 1[/math] btw).

[Johnny5]

Looks fine to me. That's certainly we way I derive it.

 

(It works for [math]|x| < 1[/math] btw).

 

Well here's what bothers me...

 

Consider this please:

 

[math] \frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{9}{10000}+...+\frac{9}{10^n}+... [/math]

 

Factor out 9/10' date=' using the distributive law of algebra, so that you have:

 

[math'] \frac{9}{10} \sum_{n=0}^{n=\infty} x^n = \frac{9}{10}(1+x+x^2+x^3+...) [/math]

 

Where x = 1/10.

 

So here we have a case where |X|<1, and this is a geometric series, so that we can replace the infinite series by S, like so:

 

[math] \frac{9}{10} \sum_{n=0}^{n=\infty} x^n = \frac{9}{10}S[/math]

 

 

Where [math] S= \frac{1}{1-x} = \frac{1}{1-1/10} = \frac{1}{10/10-1/10}= \frac{1}{9/10} = \frac{10}{9} [/math]

 

Therefore:

 

[math] \frac{9}{10} \sum_{n=0}^{n=\infty} (\frac{1}{10})^n = \frac{9}{10} \cdot \frac{10}{9} =1 [/math]

 

When we (erroneously conclude that)

 

1=.999999999999999...

 

I feel there is an error in there. what do you think?

[dryga]

But 1 is equal to .999 repeating.

[Dave]

Indeed. There's already god only knows threads on this already.

[matt grime]

Oooh, Johnny gets to learn about diadic expansions, christ this is going to be tedious. Still, it comes as no shock to know he thinks real numbers and decimal expansions are the same thing.

 

Question for the credulous and stupid: why is it hard to believe there are exactly two decimal representations of a very small number of real numbers, but you're perfectly happy to accept that there are an INFINITE number of representations of the form a/b for EVERY reational number? The mind boggles, it really does.

 

The discerning reader may have realized I've just snapped in the "wtf is Johnny5 prattling on about now" sense.