Exploring the Convergence of 0.999... and the Concept of Infinity

[Zurtex]

You have not defined the original infinity so how can you say that? Nor have you defined a relationship, for example would:

$$\infty_d - \infty_c = a$$

Where a is some finite number and according to your system it would be a specific finite value, what would it be? Even if it is not a finite value according to your system it would be a specific infinite number, what would it be? I am of course assuming a > 0 under the system you have stated.

So I wonder would:

$$\frac{1}{{\infty_d}^{\infty_c}} = 0.\overline{0}n ?$$

Where n is some finite digit? Or would this actually be 0?

Furthermore your notation is flawed, you are only going to have a finite amount of infinites where as I can think of an infinite number of ways to increase the value of a number. That's all the questions I can write down at the moment as I have to go, have at least another few dozen questions to see how your system works.

 

[matt grime]

You ought to at least understand that hyper-real, or infinitesimal numbers, are not real numbers and so attempting tp cite them in defence of your ramblings of ignorance about the real numbers won't win you any arguments.

 

[Hurkyl]

CHAPTER 3

Proof #1

In this proof they call upon a previously existing summation theorem that assumes to be correct and then proceed to apply the variables to come out with the conclusion. NO ONE can accurately sum any series to infinity to begin with. This entire proof stems from that fallacy.

Would you care to explain what you mean? Do recall that the DEFINITION of the sum of an infinite series is:

$$\sum_{i=0}^{\infty} a_i = \lim_{m \rightarrow \infty} \sum_{i=0}^{m} a_i$$

As you correctly point out, one cannot directly add an infinite number of terms... which is why we're not silly enough to do it that way.


Proof #2

on the third line. x or .9999... is subtracted from both sides and the assumption is made that you will get an even number of 9 as the result.

Actually, it is fairly easy to do this in a perfectly rigorous way.. In the way mathematicians define things, the position of each digit in a decimal number is labelled by an integer.

To give a hint of the flavor of how things are done done, I'll ask you this question:

In what position does 0.999~ have a '9' that doesn't appear in 9.999~?

(again, let me remind you; by the mathematical definition, that position has to be an integer... and no integers are infinite)

Proof #3

something INFINITELY SMALL is never NOTHING.

Incorrect; zero is the only infinitessimal real number. FYI, the definition of "infinitessimal" is:

x is an infinitessimal if and only if |x| is smaller than any positive real number.

Since the decimal numbers are constructed so that they are a model of the real numbers, it follows that zero is the only infinitessimal decimal number.


Proof #4

Go ahead and do the division for yourself. One divided by three. .33333333~ forever. Where is it wrong?

It's not.

well three does not divide evenly into one no matter how many digits you take it to, so that remainder of 1 will always exist just as the digits will never terminate.

But, the way mathematicians define the decimals, they don't have remainders. So 1/3 is 0.333~.

so 1/3 would equal 0.33~ R1/3

So this begs the question, in your system, what is 1/3? We can keep substituting:

1/3 = 0.33~ R1/3 = 0.33~ R0.33~ R 1/3 = 0.33~ R0.33~ R 0.33~ R 1/3

but we'll never come to a satisfactory representation.

But consider a much more pressing issue; how can a remainder possibly make sense if we want a decimal representatino of √2, or pi?



Also, allow me to suggest something for "fun". Let's assume for a moment that you actually have a well-defined and consistent number system.


Allow me to invent a new number system by decreeing that any two numbers that have the same decimal part are equal (even if they have different remainders), and furthermore I decree true any statement that can be proved from this decree. (So, for instance, 0.999~ = 1 because you've been able to prove 0.999~ R3/3 = 1 R0) I wonder what properties this number system has?

 

[Hurkyl]

Hrm.

4/3 = 4 * 0.333~ R(1/3) = 1.333~ R(4/3)
4/3 = 1 + 1/3 = 1 + 0.333~ R(1/3) = 1.333~ R(1/3)

Problem...

 

[ram2048]

$$\frac{1}{{\infty_d}^{\infty_c}} = 0.\overline{0}n ?$$

i don't know that looks complicated. is it actually a real question or are you coming up with new inventive ways to waste my time? :D

CHAPTER 3

Proof #1

Would you care to explain what you mean? Do recall that the DEFINITION of the sum of an infinite series is:

$$\sum_{i=0}^{\infty} a_i = \lim_{m \rightarrow \infty} \sum_{i=0}^{m} a_i$$

As you correctly point out, one cannot directly add an infinite number of terms... which is why we're not silly enough to do it that way.


Proof #2



Actually, it is fairly easy to do this in a perfectly rigorous way.. In the way mathematicians define things, the position of each digit in a decimal number is labelled by an integer.

To give a hint of the flavor of how things are done done, I'll ask you this question:

In what position does 0.999~ have a '9' that doesn't appear in 9.999~?

shifting the decimal for such a number is tricky business because it causes an inequality in the infinities in both expressions.

(again, let me remind you; by the mathematical definition, that position has to be an integer... and no integers are infinite)

Proof #3

i'm kinda going through this with someone in another thread at the moment as well and there is no satisfactory answer yet thus far. but expression 1 would have infinite 9's and expression 2 would have infinite+1 9's.

Incorrect; zero is the only infinitessimal real number. FYI, the definition of "infinitessimal" is:

x is an infinitessimal if and only if |x| is smaller than any positive real number.

Since the decimal numbers are constructed so that they are a model of the real numbers, it follows that zero is the only infinitessimal decimal number.


But, the way mathematicians define the decimals, they don't have remainders. So 1/3 is 0.333~.

So this begs the question, in your system, what is 1/3? We can keep substituting:

1/3 = 0.33~ R1/3 = 0.33~ R0.33~ R 1/3 = 0.33~ R0.33~ R 0.33~ R 1/3

but we'll never come to a satisfactory representation.

But consider a much more pressing issue; how can a remainder possibly make sense if we want a decimal representatino of √2, or pi?

the remainder is only used to create rational decimals for rational fraction values that would otherwise create non-terminating irrationals

the r(1/3) is not at the end of the series, rather it is part of the last digit. so you wouldn't be able to string it like that.

you could consider the last digit to be 10/3 rather than 3 <basically>

Also, allow me to suggest something for "fun". Let's assume for a moment that you actually have a well-defined and consistent number system.


Allow me to invent a new number system by decreeing that any two numbers that have the same decimal part are equal (even if they have different remainders), and furthermore I decree true any statement that can be proved from this decree. (So, for instance, 0.999~ = 1 because you've been able to prove 0.999~ R3/3 = 1 R0) I wonder what properties this number system has?

sounds like you're describing the current number system...

Hrm.

4/3 = 4 * 0.333~ R(1/3) = 1.333~ R(4/3)
4/3 = 1 + 1/3 = 1 + 0.333~ R(1/3) = 1.333~ R(1/3)

Problem...

you messed it up :D

4/3 = 4 * 0.333~ R(1/3) = 1.333~ R(1/3)
3x.333~ made 999~ the 3/3 completed the 1 which was part of the last digit 9. creating 10/10 that completes all the way to the top to make 1.

the process remainder is part of the last digit not after it.

 

[Integral]

3x.333~ made 999~ the 3/3 completed the 1 which was part of the last digit 9.

You now need to share with us what you mean by a digit. How is something "part of a digit"

 

[ram2048]

sorry if it wasn't clear enough.

when you divide out 1/3 each digit is created by a process of 10/3. each digit = 9/3 however. that 1/3 left over is the remainder I'm talking about.

 

[Integral]

You are correct, it is not clear. What do you mean when you use the word "digit"?

 

[ram2048]

suppose i wanted to get the true decimal value of 1/3 (one divided by three)

i would start doing division (the tedious way)

[pre]
0.333~ r 1
3|1.0
9
10
9
10
[/pre]

each "digit" of 3 in the decimal notation resulting was created by dividing 10 by 3.

to use hurkyl's notation that he brought up
f(n)=0 where n ≥ 1
f(n)=3 where n < 1

those digits.

except in this case

f(n)=(10/3) where n = ∞

hope that formatting works

 

[Integral]

Ok, how is something "part" of a digit? Is not a digit a single bit of information how am I to interpret something as part of a single piece?

You have shown me how you generate digits but you still have not told me what a digit is.

 

[ram2048]

okay... a digit(in base ten) is one integer of values 0 through 9 which exist in sequence to describe real numbers such that these digits multiplied by various powers of 10 can be summed to create the number they describe.

i'm going to assume that your follow-up question will be "if you just said the digit is integers 0 through 9 how can you have a digit that is 10/3"

the answer is, this is new notation

 

[Hurkyl]

shifting the decimal for such a number is tricky business because it causes an inequality in the infinities in both expressions.

Can you state yourself in a precise way?

Here are two interesting questions for your system:

What is 9 + 0.99~?
What is 0.99~ * 10?

Can you even divine a difference in your system?

In the standard mathematical definition of decimals, the two are the same string of digits (and thus equal as decimals), because for every integer n, the digit in the n-th place in the first is equal to the digit in the n-th place in the second.

i'm kinda going through this with someone in another thread at the moment as well and there is no satisfactory answer yet thus far. but expression 1 would have infinite 9's and expression 2 would have infinite+1 9's.

Mathematicians do study these kinds of things; they're called order types. There really is a way to rigorously say that a sequence has length "infinity + 1" (where "infinity" is the order type of the positive integers). However, there is no such thing as an "infinity - 1"; the problem is that in order to take things off of the end of an ordering, the ordering has to have an end.

Because the digits to the right of the decimal place are indexed by the positive integers (or negative, if you prefer), this ordering doesn't have an end from which things can be removed. This is why it is possible to left shift all of the digits and have "one more" than you started with. (Mathematically, it still has the same order type and cardinality)

You, like most others who have similar ideas, seem to be imagining an order type that does have an end. Now, if you were to define the decimals as having two groups of digits to the right of the decimal place, an "a group" and a "b group" so that the positions went like this:

| 1a, 2a, 3a, ... | ... 3b, 2b, 1b |

(where I've used pipes (|) to denote the boundary of a group)

then when you multiplied this new type of decimal by 10, you would indeed "lose" a digit off the right end; the 1b position would be filled with a zero.

sounds like you're describing the current number system...

Exactly right. I was trying to demonstrate that even if you did have a consistent number system with infinitessimals, you can recover a number system that behaves exactly like the standard mathematical system. The point is to show relative consistency; if you believe the standard system to be flawed, then your system must be flawed as well because your system can be used to create the standard system.

 

[Hurkyl]

Here's a puzzler. If you are only introducing remainders in the case of dividing things, then what is e^(-ln 3)?

 

[Integral]

okay... a digit(in base ten) is one integer of values 0 through 9 which exist in sequence to describe real numbers such that these digits multiplied by various powers of 10 can be summed to create the number they describe.

i'm going to assume that your follow-up question will be "if you just said the digit is integers 0 through 9 how can you have a digit that is 10/3"

the answer is, this is new notation

So once again what do you mean by part of a digit? You are going to need a better definition then simply saying it is notation. The ellipsis is simply notation, I can define it in a few words; it stands for a infinitely repeating pattern. Now what do you mean by "part of a digit" ?

You seem to have agreed that infinity must have some default value, which to me says that you have confused infinity with a large finite number, all of your work is consistent with that conclusion. If you replaced infinity with any large finite number all of your work would make sense and you would have little disagreement from anyone on this board.

Your belief that 1 - .999... >0 means that the interval (.999...,1) has a finite non zero length, thus there must be a hole in the real line at this point, are you saying that .999... is the hole? are there other holes? If there is a hole at .333... why is it that .1(base 3) exists is it not right in the middle of the hole that you have placed at .333...?

I think it is your system that is full of holes, it simply does not work. Your time would be better spent actually attempting to learn how the real number system works rather then trying to reinvent a square wheel. It's your time.

 

[Hurkyl]

Consider this.

Choose any of the 9's in 0.999~.
Consider the 9 in the next position to the right.
This 9 gets shifted underneath the 9 you chose when you multiply by 10.
Thus, 9.999~ has a 9 in the same position as the 9 you chose in 0.999~.

If this argument holds for any 9 you choose, then we have proved that 0.999~ cannot have a 9 in a position that 9.999~ does not.


Again, this works because decimal numbers don't have right ends; each position has a next position to the right.

 

[Zurtex]

Not trying to waste your time at all, just showing you that your system does not work at all ever. But let's go back to basics, you agreed that:

$$0.3\overline{3} = \sum_{n=1}^{\infty} \frac{3}{10^n}$$

Well sorry to say but if you work that out using simple sums to infinity of geometric seris you get:

$$\sum_{n=1}^{\infty} \frac{3}{10^n} = \frac{1}{3}$$

 

[ram2048]

The point is to show relative consistency; if you believe the standard system to be flawed, then your system must be flawed as well because your system can be used to create the standard system.

not if the flaw is inherent in the creation of the standard. eg: rounding processes.

Here's a puzzler. If you are only introducing remainders in the case of dividing things, then what is e^(-ln 3)?

i have no idea what that is. please explain -ln

has a finite non zero length, thus there must be a hole in the real line at this point, are you saying that .999... is the hole? are there other holes? If there is a hole at .333... why is it that .1(base 3) exists is it not right in the middle of the hole that you have placed at .333...?

i'm saying that because these numbers have infinity included in their nature the gaps between them are beyond human comprehension and can only be measured or described by "infinitessimal" notations.

Choose any of the 9's in 0.999~.
Consider the 9 in the next position to the right.
This 9 gets shifted underneath the 9 you chose when you multiply by 10.
Thus, 9.999~ has a 9 in the same position as the 9 you chose in 0.999~.

If this argument holds for any 9 you choose, then we have proved that 0.999~ cannot have a 9 in a position that 9.999~ does not.

yes but consider a static infinity.

let's say you have two numbers of .999. Each with $$\infty_d$$
number of 9's as decimal digits.

you multiply one of them by 10.

compare the two. if they both have $$\infty_d$$ number of 9's still you have created a gap at the very end such that when you subtract the two there is not an integer result.

if you "create" another 9 (wackiness) then one has $$\infty_d$$+1(9's) and the other has $$\infty_d$$. hence they're using different infinities and the calculation is incorrect.

Well sorry to say but if you work that out using simple sums to infinity of geometric seris you get: $$\sum_{n=1}^{\infty} \frac{3}{10^n} = \frac{1}{3}$$

sums to infinity are wrong which is what I've been trying to say forever now :D

Zeno's paradox is not obliterated by "sums to infinity" within the confines of the problem set forth it creates a true statement that the destination cannot be reached

 

[russ_watters]

you should care because the traditional way of viewing it leads to inaccuracies. such that something cut infinitely results in nothing.

it's fine by me if you want to embrace something that produces the wrong results due to fallacious logic.

go ahead

Math is a tool with which we can model the real world. Its of course necessary that the capabilities of the tool exceed the requirements of any concievable application.

What you see as a flaw in math is actually just part of its power and flexibility. But the problem of course, is that in order to get the correct answers from math, you have to use it correctly:

I think it is your system that is full of holes, it simply does not work. Your time would be better spent actually attempting to learn how the real number system works rather then trying to reinvent a square wheel. It's your time.

Not only do you see holes that aren't there, you're trying to patch them with a wire mesh. Yeah - it really would be better for you if you learned not only how math works, but how to use and apply it correctly.

 

[Zurtex]

sums to infinity are wrong which is what I've been trying to say forever now :D

Then you think $0.\overline{3}$ is inherently wrong anyway and thus this discussion is pointless.

If you think sums to infinity are wrong you must then believe that we have got values such as $e$, $\pi$ and $\sqrt{2}$ wrong. Please tell use how you define such values.

 

[Hurkyl]

not if the flaw is inherent in the creation of the standard. eg: rounding processes.

It's straightforward to show that this methodology is consistent; the "danger" is that it might wind up just saying that everything is equal.

i have no idea what that is. please explain -ln

ln is the natural logarithm (LN, but lowercase)

i'm saying that because these numbers have infinity included in their nature the gaps between them are beyond human comprehension and can only be measured or described by "infinitessimal" notations.

If you can measure or describe them by infinitessimal notations, doesn't that place them in the realm of human comprehension?

Something needs clarified here; are you saying that the real numbers has holes, or just the decimal numbers (whatever you think the decimal numbers are)?

yes but consider a static infinity.

I can't; I don't know what a "static infinity" is... but from

you have created a gap at the very end

I'm assuming you mean an order type that has an end. As I stated, if you are using an order type that has an end, then this argumentat is correct. However, this argument fails when the order type has no end. Since there is no last integer, the decimals as they are mathematically defined has no end, so your argument fails.

sums to infinity are wrong which is what I've been trying to say forever now :D

Are you disagreeing with the claim that

$$\lim_{m \rightarrow \infty} \sum_{n=1}^m \frac{3}{10^n} = \frac{1}{3}$$

? Or are you disagreeing with

$$\sum_{n=1}^{\infty} \frac{3}{10^n} = \lim_{m \rightarrow \infty} \sum_{n=1}^m \frac{3}{10^n}$$

?

 

[ram2048]

disagreeing with

$$\sum_{n=1}^{\infty} \frac{3}{10^n} = \lim_{m \rightarrow \infty} \sum_{n=1}^m \frac{3}{10^n}$$

that.

the sum will never equal its limit because infinity is infinite

when you say .333 that's a sum when you say 1/3 that's its limit

i'm sure I'm not saying it correct but it should be correct enough to understand.

.333 approaches 1/3 but never reaches it.

.999 approaches 9/9 but never reaches it.

ps> i have no knowledge of how to perform logarithmic calculations and don't have a sci calc handy (not that it would help since i wouldn't know what to put into begin with..)

 

[Zurtex]

Ahh well that clears it up, you simply don't understand what infinity, a limit, a sum to infinity or a recurring decimal is.

 

[Hurkyl]

the sum will never equal its limit because infinity is infinite

Ok, then this is one precise point where we differ.


Mathematically, $\sum_{n=1}^{\infty} a_n$ means $\lim_{m \rightarrow \infty} \sum_{n=1}^m a_n$. (i.e. it's neither a theorem nor an assumption; this is how an "infinite sum" is defined)

 

[ram2048]

ah so i have to work at getting that abolished first then eh?

because it makes no logical sense that the sum of a convergent series equals it's limit when the very definition of convergent means it will never reach its limit.

so there we go.

 

[arildno]

the very definition of convergent means it will never reach its limit.

?

 

[russ_watters]

the very definition of convergent means it will never reach its limit.

so there we go.

This is another definition on which you are wrong (or 'don't agree with' - which in this case is pretty much the same thing). "Convergent" means it does reach its limit. "Divergent" means it doesn't.

 

[ram2048]

Convergent: - Mathematics. The property or manner of approaching a limit, such as a point, line, function, or value.

hmm Wolfram Mathworld studiously avoids defining convergent...

possibly another conspiracy... ;D

 

[Hurkyl]

A sequence is convergent iff it has a limit.


BTW, have you considered the sequence: 0, 0, 0, 0, ...?


I think what you meant to say is that, generally, each term of the sequence will be inequal to the limit. (But, of course, that is not always true)

because it makes no logical sense that the sum of a convergent series equals it's limit when the very definition of convergent means it will never reach its limit.

I think what you meant to say is that it violates your "common sense".

Even with your definitions this doesn't follow; even if the partial sums don't "reach" the limit, why should that suggest anything about the infinite sum?

 

[ram2048]

for fun's sake let's go back to Zeno's Paradox

1/2 + 1/4 + 1/8 + 1/16...

every step only completes half of the remaining value converging towards 1.

BUT let us consider the last digit

1/$$\infty$$

here's where we get the problem i think

1/$$\infty$$ would complete only half distance theoretically as well, BUT current math doesn't believe beyond infinity.

AND

they believe that anything divided by infinity is the same number

hence you COULD have 2/$$\infty$$ and complete the whole remaining distance and it would be the same as adding 1/$$\infty$$

pure conjecture of course. i think that in order to continue our understanding of math we have to find ways to define stuff better. having 20 different definitions for infinity is a pain in the patoot

 

[Integral]

having 20 different definitions for infinity is a pain in the patoot

I sure understand what you are saying, that is why I cannot figure out why you insist on having so many definitions of infinity.

If you would simply make an effort to understand the SINGLE definition used by Mathematicians your world would be much simpler.

In the extended real line infinity is simply defined as being larger then all real numbers. A single simple definition. Thus any real number times infinity is still infinity because it is still larger then any real number. Likewise 1 over infinity must be smaller (in absolute value) then all real numbers, the only thing that meets this condition is 0, therefore $$\frac 1 \infty = 0$$ simple and consistent. You only add unnecessary complexity with your infinite number of infinites.

You are right that a set of infinitely shrinking intervals does indeed contain something (I think you phrased it that no matter how much you divide it up there is still something there). That something is a single point. Will you agree that the "length" of a single point is 0? That is 1-1=0? So does it not make sense to say that the length of the interval resulting in infinite divisions (which results in a single point) has length zero?

Will wait for your reply to make sure we are on the same page.

 

[matt grime]

Ram seems to be having the standard issue that sequences can tend to limits and not actually reach them after a finite number of steps. That aand the facthe thinks there's a last term in an infinite sequence.

Would you like to list the 20 different definitions of infinity? There's only one meaning for infinite, and no need to use the word infinity if it confuses you.

 

[ram2048]

In the extended real line infinity is simply defined as being larger then all real numbers. A single simple definition. Thus any real number times infinity is still infinity because it is still larger then any real number.

so you consider infinity to be a number governed by its own rules such that infinity is greater than itself (greater than all real numbers)

or is it not a number, such that it doesn't break that definition being greater than itself should infinity + 1 > infinity

and as far as the interval being a single point i don't believe it so. I still believe everything is further infinitely divisible beyond infinity.

 

[arildno]

ram_1024:
(I won't be part of this ego-doubling process from your side; however, I'm not a churl, so I won't start halving it either)

I'd like to point out a few implications of your own ideas, rather than making clear to you why standard maths is consistent, and hence, why your attacks on it is quixotic at best.

(Others on this forum are by far more competent than myself in doing such an explanation, and if you had bothered to read, and tried to digest what they have patiently written to you, you would have stopped these ridiculous attacks long ago)

Implications of your ideas:
a)
Now, you say that a sequence: 0.1, 0.01, 0.001 and so on does not converge to 0,
but to some number 0...1 (with an infinite number of zeroes in between)
Clearly then, a sequence: 0.2, 0.02, 0.002 goes to some number:
0...2 (Right?)

In particular, this is perfectly in accordance that each term in the second sequence is the double value of the same term in the first sequence:
For example, 0.02=2*0.01, so it should make sense that the end value,
0...2 is the double of 0...1 (Correct?)

b) Look now at the sequence:
1, 0.1, 0.01 and so on.
At every single instance, the term in this sequence is 10 times bigger than the same term in the sequence 0.1, 0.01 and so on.

Hence, by your logic, 1, 0.1, 0.01 must go to the number 0...10
(You can't escape this conclusion, sorry about that!)

c)
However, the first sequence is simply a subsequence of the last one; i.e. subsequences in your system doesn't go to/converge to the same values as the sequence itself.
In short, the whole convergence concept is blown to smithereens, and it is meaningless in the first instance to say that something "produces" or goes to
something at all; i.e. every single utterance you have made is mumbo-jumbo and nothing else.

Have a good day!

 

[matt grime]

there is nothing contradictory about this infinity+1 not being greater than infinity, since the definition is that it is greater than all real numbers; infinity is not a real number, so infinity+1 is still infinity, unless you're talking ordinals, when w and w+1 are distinct ordinals, but that is a different system from the extended real line again. amazingly it always seems that these crackpot attempts at showing inconsistency are inconsistent, but also that someone non-crackpot has thought it through and offered something that is consistent and does it properly: surreal numbers, hyperreal numbers, the extended complex plane, ordinals, cardinals...

 

[Hurkyl]

Suppose I do a thorough analysis of red M&Ms. Does that mean there are no green M&Ms?

The destination, indeed, does not appear in Zeno's analysis, but there's no reason in particular to think that Zeno's analysis covers the entirety of the motion in question.

However, there is not a "remaining distance" either. Zeno's analysis covers every position up to (but not including) the destination. If the destination is 1 meter away, then Zeno's analysis covers all positions x where 0 <= x < 1. The proof of this requires the Archmedian property posessed by the real numbers: for every real number r there is an integer n such that n > r.

It goes roughly as follows: let x be any real number in 0 <= x < 1. Let y = 1/(1-x). By the archmedian property, there is an integer n > y. By induction, we can find an integer m with 2^m > n. Thus, there is an integer m with 2^m > y, so 1/(2^m) < 1-x and 1 - 1/(2^m) > x. However, 1 - 1/(2^m) is the "current position" of the runner after the m-th step of Zeno's analysis. Thus, the position x has been considered by Zeno.

1/2 + 1/4 + 1/8 + 1/16...

every step only completes half of the remaining value converging towards 1.

BUT let us consider the last digit

Did you mean last term?

And why would there be a last term? Each term is of the form 1/2^m where m is an integer and there is no last integer...

so you consider infinity to be a number governed by its own rules such that infinity is greater than itself (greater than all real numbers)

The system Integral describes is called the "extended real numbers". In this system, infinity and -infinity are extended real numbers that are not real numbers.