I really must know the answer to this important mathematical question

Because no number in the series added to any number in the series equals one, either. It isn't that kind of series.

.9 + .99, for example, equals 1.89. That is far more than one.

.9999 + .99999 = 1.99989

.999999999999 + .9999999999999 = 1.9999999999989

 

Wrong series. You want

.9 + .09 + .009 + .0009...

Which as Kubo said is the just the correct way to write .9999...

My proof that I've been building to is just solving that series.

 

Aaah, you're right.

But still, it will never equal one. Though I concede my logic wasn't sound in itself. I didn't understand the nature of a series before this.

Is there anything else I should know?

 

Actually I just noticed a way to do this without limits. I was stuck on n starting at 1 but I can easily change that to 0.

Σ(0.9*(0.1)^n) where n starts at 0 and goes to infinity. This is a slightly different way to write the same series as above.

This is a geometric series since any term after the first is a constant factor (1/10) of the previous term. And it converges because our r of 0.1 has an absolute value less than 1.

So we can use the formula s=a/(1-r) where a=0.9 and r=.1
Thus s=.9/.9
So s=1

 

Okay, I follow you up to here.

But I don't understand why that formula exists or what it means, and thus why I should accept it. Please explain in more detail. Also what the r is.

Because it seems to me that you're saying n = n / n = 1

But then why couldn't I argue that .3333... / .3333... = 1?

It feels like a nonsense equation.

 

When I say you lack basic calclulus knowledge, you think that I say it because I don't know what to tell you. This formula is used in geometric series that converge and has a proof.
r is the common ratio; 1/10.
If you want to see what 0.333 does then you need to go to the formula and see what happens. Not every series =1. In fact this series converge to 3/7, if I did the calculations correctly.

 

We'll there is an algebraic proof of the geometric formula.
The formula applies to a series of the form

Σ(a*r^n) where n starts at 0 and goes to infinity.

So s = a + ar + ar^2 + ar^3....
s = a * (1 + r + r^2 + r^3....)

Now lets break that apart for a moment.
s = a*u
u = 1 + r + r^2 + r^3....

u*r = r + r^2 + r^3 + r^4....
u - u*r = 1
u(1-r) = 1
u = 1/(1 - r)

Then substitute u back in
s = a (1/(1 -r))
s = a/(1 - r)

 

I don't assume that. I am however trying to provoke you into telling me by pretending as if I assume that.

Okay, so can you explain to me the reason for the geometric formula (what it's output is supposed to be) and why I should accept it? My fundamental questions were not about the proof, but about why said proof discounts my logic.

Oh, and I don't get this:

u - u*r = 1

Where did the 1 come from? Or rather, back here:

u*r = r + r^2 + r^3 + r^4....

Where did it go and why did it come back?

 

u = 1 + r + r^2 + r^3.... (1)
u*r = r + r^2 + r^3 + r^4.... (2)

(1)-(2)=u-u*r= (1 + r + r^2 + r^3....) - ( r + r^2 + r^3 + r^4....) = 1

 

What?, I hope you're happy. This is entirely your fault.

 

... But aren't you already assuming that the series 1 + 1 + r + r^2 + r^3... is equal to one? Why does it equal one?

And you didn't explain where the 1 went to, or how it suddenly gained a number even thought it lost the number 1. As far as I can tell, you are saying:

u = 1 + r + r^2 + r^3....

But that:

u * r = u - 1

 

I suggest you let it go because no matter what me or someone else tells you it will take a long time before you understand it without experience. I don't really have the urge to explain to you anything else.

 

I think this is something even I am capable of elucidating.

The 1 didn't go anywhere, it was multiplied by r. In detail:

u = 1 + r + r^2 + r^3 +r^4...

u * r = r(1) + r(r) + r(r^2) + r(r^3) + r(r^4)... = r + r^2 + r^3 + r^4 + r^5...

Now, about the sequences' difference, do you agree that every power of r in u is also present in u * r? More clearly:

u = 1 + (powers of r); u * r = (powers of r)

In the end, subracting u * r from u results in 1.

 

Bingo.

You have to put the series out of phase, but addition is commutative so it doesn't matter. This phase offset would distort the end result slightly when you run out of "r"s in a limited summation, but in an endless summation saying that one summation ends one term before the other doesn't make sense, neither has an end and thus one can't end before the other.