1=0.999... infinities and box of chocolates..Phliosophy of Math...

[Baldeee]

What point you are trying to prove here?

You previously wrote:

$$T_n$$ can only tend towards $$0$$ but $$T_n$$ can not be equal to $$0$$.

I have shown that, if you consider $$T_\infty$$ part of the geometric sequence then this must be zero.
i.e. if we conceptualise an end to the infinite sequence, then it must be zero.

If you don't include it as part of the sequence (as rpenner suggests to be the case) then that would be another matter.

 

[hansda]

You previously wrote:

I have shown that, if you consider $$T_\infty$$ part of the geometric sequence then this must be zero.

What about the constant ratio between the successive terms there?

i.e. if we conceptualise an end to the infinite sequence, then it must be zero.

Infinite means not finite or no end.

If you don't include it as part of the sequence (as rpenner suggests to be the case) then that would be another matter.

Do you think it will change from geometric series to some other series?

 

[hansda]

Only in math systems where $$\infty + 1 = \infty$$ like the extended real line. In such number systems $$0 = \frac{9}{\infty} = \frac{9}{10^{\infty}} = 9 \times 10^{-\infty}$$ But it is still bending the rules since the family of $$T_n$$ was only defined for n in the counting numbers. Since $$\infty$$ is not a counting number, this $$T_{\infty}$$ is still not part of the family being considered.

So just in number systems where $$\infty + 1 = \infty$$, we can correctly write $$\lim_{n\to \infty} S_n = S_{\infty} = S = 1$$ while in the ordinary reals or hyperreals we can write $$\lim_{n\to \infty} S_n = S = 1$$ and still be correct.

What about the constant ratio between successive terms of a geometric series? Will it still hold when $$T_\infty = 0$$?

The constant ratio is $$\frac{T_1}{T_2} = \frac{T_2}{T_3} =...= \frac{T_(n-1)}{T_n}=\frac{T_n}{T_(n+1)}=...=\frac{T_(\infty-1)}{T_\infty}=\frac{T_\infty}{T_(\infty+1)}$$.

 

[BdS]

@Bds,
My question:

Your response:

This just beggars the question:

What is the number 1?
From what I understand the number 1 is the sum of it's infinite constituents.
so therefore the sum of 1 = 0.999.... [and all 0.999... means]
(where the value 1 becomes the upper boundary of 0.999... perhaps )

*edit: I think this was cleared up later in the thread [still reading]

I'm in way over my head in this thread... I'd prefer not to say anything more in this thread, but my answer to you should have probably been yes. Most posts in the thread are beyond my abilities with maths and I dont want to bring more of my confusion in here. After rpenner taught me that base 10 goes from 0 - 9, it showed me I dont even know the basics and my posts in maths arguments are not worthy to be considered or even read by others. I will just keep reading/learning and try to understand and listen to those with a formal education on such matters.

 

[hansda]

What about the constant ratio between successive terms of a geometric series? Will it still hold when $$T_\infty = 0$$?

The constant ratio is $$\frac{T_1}{T_2} = \frac{T_2}{T_3} =...= \frac{T_(n-1)}{T_n}=\frac{T_n}{T_(n+1)}=...=\frac{T_(\infty-1)}{T_\infty}=\frac{T_\infty}{T_(\infty+1)}$$.

From the above we can safely say that, $$T_n$$ can never be $$0$$ for any value of $$n$$. So, we can write $$T_n \ne 0$$; for any value of$$ n$$.

Another thing that is true is for every counting number, n, is "$$S_n + \frac{1}{9} T_n = 1$$."

Considering the above equation we can write, $$S_n = 1- \frac{1}{9} \times T_n$$.

As $$T_n \ne 0$$for any value of $$n$$, we can say that $$S_n \ne 1$$ for any value of $$n$$.

So, we can say $$\lim_{n \to \infty} {S_n} = S = 0.999... \ne 1$$.

 

[hansda]

Have you found a flaw in post #915?



Ok. Please write one down for the benefit of Motor Daddy, hansda and the others. I'd like to see your simple proof, too.

I think I observed a flaw in the post #915. See my post #1065.

Let's consider a different example for a minute.

What is 1/7?

Plug it into your calculator and you'll get an answer like 0.1428571429.

That's rounded, though. The actual answer is:

0.142857 142857 142857 ...

where the pattern repeats to infinity.

I am wondering whether Motor Daddy, hansda or others think that what I have said here is actually wrong. That is, do you think that pattern stops at some point? If so, what is the last digit in the decimal expansion of 1/7?

Once you've answered that, I want you to multiply the decimal number by 7 and tell me what you get as a decimal.

In particular, what is the last digit in the result of 7 x 0.142857142857...etc.?

See, because the problem I have is that if I say the last digit is, say, 2, then when I multiply the whole thing by 7 then the last digit should be a 4. And yet, the answer should be 7.000000....

So, Motor Daddy, hansda and others, what's going wrong in this case?

Here also we can say that $$\frac{1}{7} \ne 0.142857 142857 142857 ...$$

And if that one's too hard, consider this:

What is 1/9, expressed as a decimal?

I say it's 0.1111...

If you disagree, then what do you say it is?

And what happens when I do this:

0.111... x 9 = ?

It seems we can agree that 0.111... x 9 = 0.999..., but if 0.999... doesn't equal 1, then something's gone horribly wrong somewhere.

So, explain it to me.

I already proved that $$1 \ne 0.999...$$.

So, we also can say that $$\frac{1}{9} \ne 0.111...$$

 

[arfa brane]

So, we can say $$\lim_{n \to \infty} {S_n} = S = 0.999... \ne 1$$.

Sure, you can say lots of things; in this case the above is a mistake, it's wrong, it does not follow from the rest of your "argument".

I already proved that $$1 \ne 0.999...$$.

So, we also can say that $$\frac{1}{9} \ne 0.111...$$

You didn't prove that $$1 \ne 0.999...$$, you just stated it as if it was true.

Have you read and understood any of the proofs posted by others? No you haven't? That's a shame.

 

[rpenner]

[post=3164396]Post #915[/post] 2014-02-23

Hansda, consider $$0.999... = \sum_{k\geq 1} \frac{9}{10^k}$$
... $$S_n = \frac{9}{10} \times \frac{1 - 10^{-n}}{1 - 10^{-1}}$$ is true for all counting numbers, n.
But because $$\frac{9}{10} = 1 - 10^{-1}$$ this also proves that $$S_n = 1 - 10^{-n}$$ for all counting numbers, $$n$$.

[post=3165524]Post #990[/post] 2014-02-26

Another thing that is true is for every counting number, n, is "$$S_n + \frac{1}{9} T_n = 1$$."

In summary:

$$S_n = 1 - 10^{-n}, \quad T_n = 9 \times 10^{-n}, \quad \lim_{n\to\infty} S_n = S = 0.999... , \quad \lim_{n\to\infty} T_n = T \\ \forall n \in \mathbb{N}^+ \quad S_n + \frac{1}{9} T_n = 1 \; \wedge \; S_n \lt S \; \wedge \; T_n \gt T \\ S \not\in \left{ S_n \; | \; n \in \mathbb{N}^+ \right} \\ T \not\in \left{ T_n \; | \; n \in \mathbb{N}^+ \right} \\ x \lt S \quad \Rightarrow \quad \forall n \in \mathbb{N}^+ \quad n \gt - \log_{10} ( 1 - x ) \; \rightarrow \; x \lt S_n \lt S \\ S = 1 \\ x \gt T \quad \Rightarrow \quad \forall n \in \mathbb{N}^+ \quad n \gt - \log_{10} ( x / 9 ) \; \rightarrow \; T \lt T_n \lt x \\ T = 0 \\ \forall k \in \mathbb{N}^+ \quad \textrm{card} \left{ n \, | \, n \in \mathbb{N}^+ \; \wedge \; \neg n \gt k \right} \quad = \quad k \\ \forall k \in \mathbb{N}^+ \quad \left{ n \, | \, n \in \mathbb{N}^+ \; \wedge \; \neg n \gt k \right} \quad \prec \quad \left{ n \, | \, n \in \mathbb{N}^+ \; \wedge \; n \gt k \right} \\ \infty \not\in \mathbb{N}^+ $$​

[post=3167065]Post #1059[/post] 2014-03-02

May be you are right but do you think $$T_n$$ being a valid member of a geometric series can ever be $$0$$ for any value of $$n$$?

[post=3167067]Post #1060[/post] 2014-03-02

Only in math systems where $$\infty + 1 = \infty$$ like the extended real line. In such number systems $$0 = \frac{9}{\infty} = \frac{9}{10^{\infty}} = 9 \times 10^{-\infty}$$ But it is still bending the rules since the family of $$T_n$$ was only defined for n in the counting numbers. Since $$\infty$$ is not a counting number, this $$T_{\infty}$$ is still not part of the family being considered.

So just in number systems where $$\infty + 1 = \infty$$, we can correctly write $$\lim_{n\to \infty} S_n = S_{\infty} = S = 1$$ while in the ordinary reals or hyperreals we can write $$\lim_{n\to \infty} S_n = S = 1$$ and still be correct.

[post=3167081]Post #1063[/post] 2014-03-02

What about the constant ratio between successive terms of a geometric series? Will it still hold when $$T_\infty = 0$$?

In the case where $$T_{\infty} = 0$$ then as I have already said that only happens in number systems where $$\infty + 1 = \infty$$, and thus $$\frac{T_{\infty-1}}{T_{\infty}} \equiv \frac{T_{\infty}}{T_{\infty}} $$ and $$\frac{T_{\infty}}{T_{\infty+1}} \equiv \frac{T_{\infty}}{T_{\infty}}$$ are equivalent to $$\frac{0}{0}$$ and are undefined. This doesn't break the assumption of geometric series because when $$\infty$$ doesn't obey the axioms of counting numbers then there are no successive terms relative to $$T_{\infty}$$.

The constant ratio is $$\frac{T_1}{T_2} = \frac{T_2}{T_3} =...= \frac{T_(n-1)}{T_n}=\frac{T_n}{T_(n+1)}=...=\frac{T_(\infty-1)}{T_\infty}=\frac{T_\infty}{T_(\infty+1)}$$.

As I point out, you have bad math in this line.
[post=3167333]Post #1065[/post] 2014-03-03

From the above we can safely say that, $$T_n$$ can never be $$0$$ for any value of $$n$$. So, we can write $$T_n \ne 0$$; for any value of$$ n$$.

True and irrelevant if you had understood post #915. n is restricted to being a counting number in all of post #915 and the conclusion that $$\forall n \in \mathbb{N}^+ \; T_n \gt T$$ is true.

There is no counting number that makes $$S_n =^? S$$ or $$T_n =^? T$$ true. In a limited sense, the $$\infty$$ of the extended real line makes $$S_{\infty} = S$$ and $$T_{\infty} = T$$ true, but that does not apply to the hyperreals as it explicitly depended on the property of the $$\infty$$ of the extended real line such that $$\infty \pm 1 = \infty$$. However, in the counting numbers, in the extended real line and in the hyperreals, $$\lim{n \to \infty} S_n = S$$ and $$\lim{n \to \infty} T_n = T$$.

Considering the above equation we can write, $$S_n = 1- \frac{1}{9} \times T_n$$.

True. Importantly you can also write $$S = 1- \frac{1}{9} \times T$$.

As $$T_n \ne 0$$for any value of $$n$$, we can say that $$S_n \ne 1$$ for any value of $$n$$.

This is still true, but irrelevant because $$S_n \ne S$$ when n is any counting number.

So, we can say $$\lim_{n \to \infty} {S_n} = S = 0.999... \ne 1$$.

The symbol $$\neq$$ does not follow from any line of argument. It is simply nakedly asserted.
[post=3167862]Post #1066[/post] 2014-02-04

I think I observed a flaw in the post #915. See my post #[post=3167333]1065.[/post]

Not a flaw. You observed a feature of the real numbers and of limits in the real numbers and the hyperreals that you didn't understand. You also didn't understand that the argument about the infinity extended real number line was based on it having different properties both from the counting numbers and the infinitely large numbers of the hyperreals. By mixing different formal systems you have arrived at a result which is true in none of them.

 

[rpenner]

Translations of my "summary" into English.

$$S_n = 1 - 10^{-n} $$

For any counting number, $$n$$, the expression $$S_n$$ is just a shorthand way of writing the quantity $$1 - 10^{-n}$$ and this follows from the original definition of $$S_n$$ being the finitely repeating decimal 0.9...9 with $$n$$ 9's, or the same value expressed as a geometric sum which starts with 0.9 and each term is smaller by a factor of 10, or the clever formula for writing in fixed form the sum of any finite geometric series. So all of these terms are exactly equal to each other even though they don't look much alike:

$$S_n = 0.\underbrace{999...999}_{n \; \textrm{nines}} = \sum_{k=1}^n \frac{9}{10^k} = \frac{9}{10} \times \frac{1 - 10^{-n}}{1 - 10^{-1}} = 1 - 10^{-n}$$​

$$ T_n = 9 \times 10^{-n} $$

For any counting number, $$n$$, the expression $$T_n$$ is just a shorthand way of writing the quantity $$9 \times 10^{-n}$$ and since $$S_n = \sum_{k=1}^n T_k$$ it follows that $$T_{n+1} = S_{n+1} - S_n$$ and also $$S_n + \frac{1}{9} T_n = 1$$

$$\lim_{n\to\infty} S_n = S = 0.999... $$

As the parameter $$n$$ is moved to larger values, the expression $$S_n$$ moves ever closer to a unique limiting value, which we $$S$$. Conceptually, this value must be the same as if the number of repeating nines grew larger than any finite number, so we choose to also call it $$0.999...$$ which is the topic of this thread. Even though the symbol $$\infty$$ is in this expression we do not use that symbol in any formal sense. We don't set the parameter $$n$$ to infinity because that would lead us to an area where arithmetic is not defined.

$$\lim_{n\to\infty} T_n = T $$

Likewise, as the parameter $$n$$ is moved to larger values, the expression $$T_n$$ moves ever closer to a unique limiting value which we call $$T$$. If we wanted we could call that $$T = 9 \times ( 1 - 0.999... )$$ but since we don't yet know how to do the arithmetic in the inner parenthesis, I submit introducing this notation is a bad idea. We will attempt to reason out what $$0.999...$$ must be and that should solve the problem of what $$T$$ is at the same time.

$$\forall n \in \mathbb{N}^+ \quad S_n + \frac{1}{9} T_n = 1 \; \wedge \; S_n \lt S \; \wedge \; T_n \gt T $$

For any specific counting number $$n$$, all of the following are true: $$\quad S_n + \frac{1}{9} T_n = 1$$ which follows from direct calculation and $$S_n \lt S$$ and $$T_n \gt T$$ which both follow from the observation that the expression $$10^{-n}$$ is strictly decreasing for increasing values of the parameter $$n$$.

$$S \not\in \left{ S_n \; | \; n \in \mathbb{N}^+ \right} $$

The value $$S$$ is not anywhere equal to the expression $$S_n$$ for any counting number $$n$$. That makes sense since $$0.999...$$ with a greater than finite number of 9's can't equal $$0.\underbrace{999...999}_{n \; \textrm{nines}}$$ when $$n$$ is finite.

$$ T \not\in \left{ T_n \; | \; n \in \mathbb{N}^+ \right} $$

The value $$S$$ is not anywhere equal to the expression $$S_n$$ for any counting number $$n$$.

$$ x \lt S \quad \Rightarrow \quad \forall n \in \mathbb{N}^+ \quad n \gt - \log_{10} ( 1 - x ) \; \rightarrow \; x \lt S_n \lt S $$

If $$x$$ is any real number less than $$S$$ then it follows that every counting number, $$n$$, which is larger than $$- \log_{10} ( 1 - x )$$ it follows that $$S_n$$ is strictly between $$x$$ and $$S$$.

$$ S = 1 $$

From the above, it is impossible that $$S \lt 1$$ or there would be values of $$S_n$$ larger than $$S$$ while if $$S \gt 1$$ then there would be values of $$x$$ between the largest possible values of $$S_n$$ and $$S$$. Thus $$S = 1$$ and $$0.999... = 1$$.

$$ x \gt T \quad \Rightarrow \quad \forall n \in \mathbb{N}^+ \quad n \gt - \log_{10} ( x / 9 ) \; \rightarrow \; T \lt T_n \lt x $$

If $$x$$ is any real number greater than $$T$$ then it follows that every counting number, $$n$$, which is larger than $$- \log_{10} ( x /9 )$$ it follows that $$T_n$$ is strictly between $$T$$ and $$x$$.

$$T = 0 $$

This should not be a surprise and it also follows from $$T = 9 \times ( 1 - 0.999... ) = 9 \times ( 1 - 1 ) = 9 \times 0 = 0$$

$$ \forall k \in \mathbb{N}^+ \quad \textrm{card} \left{ n \, | \, n \in \mathbb{N}^+ \; \wedge \; \neg n \gt k \right} \quad = \quad k $$

If $$k$$ is any counting number, the size of the set of counting numbers not greater than $$k$$ is equal to $$k$$. Think about it. The set of the counting number not greater than 5 is {1, 2, 3, 4, 5} and the size of that set is 5.

$$ \forall k \in \mathbb{N}^+ \quad \left{ n \, | \, n \in \mathbb{N}^+ \; \wedge \; \neg n \gt k \right} \quad \prec \quad \left{ n \, | \, n \in \mathbb{N}^+ \; \wedge \; n \gt k \right} $$

If $$k$$ is any counting number, the set of counting numbers greater than $$k$$ strictly dominates the set of counting numbers not greater than $$k$$. Or: for any counting number, $$k$$ most counting numbers are larger than it.

$$ \infty \not\in \mathbb{N}^+ $$

Infinity is not a counting number. In many number systems it is not a full-fledged number that obeys the axioms of all other numbers.

 

[hansda]

What about the constant ratio between successive terms of a geometric series? Will it still hold when $$T_\infty = 0$$?

In the case where $$T_{\infty} = 0$$ then as I have already said that only happens in number systems where $$\infty + 1 = \infty$$, and thus $$\frac{T_{\infty-1}}{T_{\infty}} \equiv \frac{T_{\infty}}{T_{\infty}} $$ and $$\frac{T_{\infty}}{T_{\infty+1}} \equiv \frac{T_{\infty}}{T_{\infty}}$$ are equivalent to $$\frac{0}{0}$$ and are undefined.

Do you think the constant ratio of the successive terms of an infinite geometric series becomes undefined $$(\frac {0}{0})$$ at the infinity, whereas it remains a definite value in the finite range?

This doesn't break the assumption of geometric series because when $$\infty$$ doesn't obey the axioms of counting numbers then there are no successive terms relative to $$T_{\infty}$$.

If there are no successive terms at the infinity, then its no more an infinite series.

 

[rpenner]

Do you think the constant ratio of the successive terms of an infinite geometric series becomes undefined $$(\frac {0}{0})$$ at the infinity, whereas it remains a definite value in the finite range?

I think a infinite geometric series is only defined over the natural numbers. (You at least partially agree with me to the extent you seem to define a geometric series as having a "first" term and then an unending succession of "next" terms with a constant ratio between a term and its "next" term. This dependence on "first" and "next" generates terms that correspond to the counting numbers.) As infinity is not a natural number it follows that substituting n with infinity means the resulting term is not a member of the sum.

However, if we agree to extend our vision to the extended reals:

If $$\infty + 1 = \infty$$ then it follows that $$T_{\infty + 1} = T_{\infty}$$ Yes? But even then $$T_{\infty + 1} = \frac{1}{10} T_{\infty}$$ can still be true but only if $$T_{\infty} = 0$$. Therefore $$\infty + 1 = \infty$$ and $$10^{-\infty} = 0$$ are logicall compatible. If you assert that $$10^{-\infty} \gt 0$$ then I point out that it cannot be the case that $$\infty + 1 = \infty$$ and therefore you are not working with numbers in the extended reals.​

And if we agree to work with the hyperreals:

$$\textrm{st} \left( H^{100} 10^{-H} \right) = 0$$ for every infinitely large H, so, by definition, $$T = \lim_{n \to \infty} T_n = \textrm{st} \left( T_H \right) = \textrm{st} \left( 9 \times 10^{-H} \right) = 0$$.​

If there are no successive terms at the infinity, then its no more an infinite series.

Not true. Because there is a term for every counting number, the series has more terms than any finite number.

 

[arfa brane]

If there are no successive terms at the infinity, then its no more an infinite series.

That really doesn't make sense; if we are "at infinity", then it must be an infinite series.

n "goes to infinity" is a direct consequence of proving that there is always a number larger than n, and so although every n is finite, the set of natural numbers is infinite in size. As rpenner puts it, an infinite series (indexed by n) must have more than a finite number of terms; "more than a finite number" is the type or class of infinity referred to here.

 

[hansda]

I think a infinite geometric series is only defined over the natural numbers. (You at least partially agree with me to the extent you seem to define a geometric series as having a "first" term and then an unending succession of "next" terms with a constant ratio between a term and its "next" term. This dependence on "first" and "next" generates terms that correspond to the counting numbers.) As infinity is not a natural number it follows that substituting n with infinity means the resulting term is not a member of the sum.

However, if we agree to extend our vision to the extended reals:

If $$\infty + 1 = \infty$$ then it follows that $$T_{\infty + 1} = T_{\infty}$$ Yes? But even then $$T_{\infty + 1} = \frac{1}{10} T_{\infty}$$ can still be true but only if $$T_{\infty} = 0$$. Therefore $$\infty + 1 = \infty$$ and $$10^{-\infty} = 0$$ are logicall compatible. If you assert that $$10^{-\infty} \gt 0$$ then I point out that it cannot be the case that $$\infty + 1 = \infty$$ and therefore you are not working with numbers in the extended reals.​

And if we agree to work with the hyperreals:

$$\textrm{st} \left( H^{100} 10^{-H} \right) = 0$$ for every infinitely large H, so, by definition, $$T = \lim_{n \to \infty} T_n = \textrm{st} \left( T_H \right) = \textrm{st} \left( 9 \times 10^{-H} \right) = 0$$.​

Do you think, $$\frac{9}{10^\infty} = 0 $$ ?

 

[rpenner]

It doesn't matter what I believe. It matters which mathematical axioms I choose to work with. Different axioms, different results. In this case $$\infty$$ as a quasi-number is only defined in one of the two systems I talk about and in that system $$T_{\infty+1} = \frac{1}{10} T_{\infty} = 0$$.

But neither of those two systems is necessary to everything in [post=3164396]Post #915[/post] being correct.

 

[hansda]

Do you think, $$\frac{9}{10^\infty} = 0 $$ ?

It doesn't matter what I believe. It matters which mathematical axioms I choose to work with. Different axioms, different results. In this case $$\infty$$ as a quasi-number is only defined in one of the two systems I talk about and in that system $$T_{\infty+1} = \frac{1}{10} T_{\infty} = 0$$.

If $$\frac{9}{10^\infty} = 0 $$, there may be some issues. Here I am highlighting some of these issues:

1) As per the definition of infinite geometric series, the infinite series has to be geometric. But here we are observing that, at the infinity this series is not geometric rather undefined as the constant ratio becomes $$(\frac{0}{0})$$. So, this series should not be called as infinite geometric series.

2) If $$\frac{9}{10^\infty} = 0 $$; multiply both the sides with $$10^\infty$$. LHS becomes $$9$$ ($$9\times \frac{10^\infty}{10^\infty}= 9$$), whereas RHS becomes $$0$$ or undefined$$(0\times \infty)$$but not $$9$$.

3)$$0.999...$$ means there are infinite numbers of digit $$9$$ in this number. Adding the digit $$9$$ at the $$n-th$$ location after the decimal point means adding the number $$\frac{9}{10^n}$$ in this series. So, at the infinity as $$n \to \infty$$, effectively not $$9$$ but $$0s$$ are being added in the series and to the number $$0.999...$$.

But neither of those two systems is necessary to everything in [post=3164396]Post #915[/post] being correct.

But you bend the rules. Ref your post #1060. I dont know if it is standard math or your own interpretation of math.

 

[arfa brane]

As per the definition of infinite geometric series, the infinite series has to be geometric. But here we are observing that, at the infinity this series is not geometric rather undefined

Are you saying there's a problem with this particular series, or with every infinite geometric series?

$$0.999...$$ means there are infinite numbers of digit $$9$$ in this number. Adding the digit $$9$$ at the $$n-th$$ location after the decimal point means adding the number $$\frac{9}{10^n}$$ in this series. So, at the infinity as $$n \to \infty$$, effectively not $$9$$ but $$0s$$ are being added in the series and to the number $$0.999...$$.

How does that follow? How do 9s become 0s at infinity?

I dont know if it is standard math or your own interpretation of math.

That about sums it up . . .

 

[rpenner]

Do you think, $$\frac{9}{10^\infty} = 0 $$ ?

It doesn't matter what I believe. It matters which mathematical axioms I choose to work with. Different axioms, different results. In this case $$\infty$$ as a quasi-number is only defined in one of the two systems I talk about and in that system $$T_{\infty+1} = \frac{1}{10} T_{\infty} = 0$$.

If $$\frac{9}{10^\infty} = 0 $$,

First of all, if "standard math" is the absolute criteria of discussion, then you must observe that $$\infty$$ is not a number in any standard math. Therefore the value of $$\frac{9}{10^{\infty}}$$ is not a question that can be answered by any "standard math" of the type called "arithmetic." However, in the extended real numbers both $$1 + \infty = \infty$$ and $$\frac{9}{10^{\infty}} = 0$$ are true and in both standard analysis and non-standard analysis based on hyperreals, the statement $$\lim_{n\to\infty} \frac{9}{10^n} = 0$$ is true. And if you aren't using one of these precise systems, then you haven't yet begun to describe what you mean by $$\frac{9}{10^{\infty}}$$.

there may be some issues. Here I am highlighting some of these issues:

1) As per the definition of infinite geometric series, the infinite series has to be geometric. But here we are observing that, at the infinity this series is not geometric rather undefined as the constant ratio becomes $$(\frac{0}{0})$$. So, this series should not be called as infinite geometric series.

WRONG. The geometric series is defined only as something with a first term and a ratio you multiply by to get to the next term. $$a_{n+1} = k a_{n}$$. It does not follow from the laws of arithmetic that you are able to recover that ratio from any two terms, just from any two non-zero terms. Also the pattern "first term" and "rule to compute the next term" is a pattern which only leads to induction over the natural numbers and $$\infty$$ cannot be reached in that manner. Thus $$\infty$$ cannot be a natural number and it isn't. Finally, for the extended real numbers where the law $$1 + \infty = \infty$$ is true, it follows that if you wanted to extended the domain of applicability of the concept of geometry series to cover terms of the format $$a_{\infty}$$ and $$a_{\infty+1}$$, then both $$a_{\infty+1} = a_{\infty}$$ AND $$a_{\infty+1} = k a_{\infty}$$ must be true, so $$ a_{\infty}$$ ( except in the case where $$k = 1$$) can only be $$-\infty$$, 0, or $$+\infty$$.

2) If $$\frac{9}{10^\infty} = 0 $$; multiply both the sides with $$10^\infty$$.

Not a sensible operation. In general terms like $$0 \times \infty$$ are explicitly disallowed in the extended number line and this line of reasoning is foreclosed by basic definitions in the hyperreals. By mixing ideas from both you combine logic and illogic to get a combination which is still contaminated by illogic which reflects your trying to run before you can walk.

3)$$0.999...$$ means there are infinite numbers of digit $$9$$ in this number.

"infinite numbers of digit 9" is probably the wrong way for someone who reasons as poorly as you do about the infinite to describe it. I think you had best think of it as an "unending sequence of digit 9." If you see any difference between these phrasings, then know you are doing it wrong.

Adding the digit $$9$$ at the $$n-th$$ location after the decimal point means adding the number $$\frac{9}{10^n}$$ in this series.

Only when n is a counting number.

So, at the infinity as $$n \to \infty$$, effectively not $$9$$ but $$0s$$ are being added in the series and to the number $$0.999...$$.

No, you have confused finite induction and transfinite induction and thus you have made a bogus argument.

But neither of those two systems is necessary to everything in [post=3164396]Post #915[/post] being correct.

But you bend the rules.

WRONG. You have never understood the rules.

Ref your post #1060. I dont know if it is standard math or your own interpretation of math.

False dilemma. It is the standard, textbook description of non-standard math, specifically the extended real number system. This was predicated on your senseless use of the symbol $$\infty$$ as if it were a natural number (aka counting number). You started talking about math beyond what you have studied and it is not my fault that you are ill-prepared to even debate the veracity of my completely textbook-equivalent answer. I also return to the subject at hand (what is S in the real number system).

Only in math systems where $$\infty + 1 = \infty$$ like the extended real line. In such number systems $$0 = \frac{9}{\infty} = \frac{9}{10^{\infty}} = 9 \times 10^{-\infty}$$ But it is still bending the rules since the family of $$T_n$$ was only defined for n in the counting numbers. Since $$\infty$$ is not a counting number, this $$T_{\infty}$$ is still not part of the family being considered.

So just in number systems where $$\infty + 1 = \infty$$, we can correctly write $$\lim_{n\to \infty} S_n = S_{\infty} = S = 1$$ while in the ordinary reals or hyperreals we can write $$\lim_{n\to \infty} S_n = S = 1$$ and still be correct.

 

[hansda]

WRONG. The geometric series is defined only as something with a first term and a ratio you multiply by to get to the next term. $$a_{n+1} = k a_{n}$$. It does not follow from the laws of arithmetic that you are able to recover that ratio from any two terms, just from any two non-zero terms. Also the pattern "first term" and "rule to compute the next term" is a pattern which only leads to induction over the natural numbers and $$\infty$$ cannot be reached in that manner. Thus $$\infty$$ cannot be a natural number and it isn't. Finally, for the extended real numbers where the law $$1 + \infty = \infty$$ is true, it follows that if you wanted to extended the domain of applicability of the concept of geometry series to cover terms of the format $$a_{\infty}$$ and $$a_{\infty+1}$$, then both $$a_{\infty+1} = a_{\infty}$$ AND $$a_{\infty+1} = k a_{\infty}$$ must be true, so $$ a_{\infty}$$ ( except in the case where $$k = 1$$) can only be $$-\infty$$, 0, or $$+\infty$$.

See the Theorem 5.5 (Ratio test) on page # 16 of infinite series by KEITH CONRAD.

From this we can say that the constant ratio of the infinite geometric series can very well be maintained in the infinity also.

Hence, $$\lim_{n \to \infty} \frac{T_{n+1}}{T_n} \ne \frac{0}{0}$$.

 

[rpenner]

See the Theorem 5.5 (Ratio test) on page # 16 of infinite series by KEITH CONRAD.

I read it, and I read it better than you.

The two basic concepts of calculus, differentiation and integration, are defined in terms of limits (Newton quotients and Riemann sums). In addition to these is a third fundamental limit process: infinite series. The label series is just another name for a sum. An infinite series is a “sum” with infinitely many terms, such as
(1.1) $$ \quad \quad \quad \quad \quad \quad \quad \quad 1 + \frac{1}{4}+ \frac{1}{9}+ \frac{1}{16} + \dots + \frac{1}{n^2} + \dots $$

(page 1)

Now we define the meaning of infinite series, such as (1.1). The basic idea is that we look at the sum of the first N terms, called a partial sum, and see what happens in the limit as $$N \; \to \; \infty$$.
Definition 2.4. Let $$a_1, a_2, a_3, . . .$$ be an infinite sequence of real numbers. The infinite series $$\sum \nolimits _{\tiny n \geq 1} a_n$$ is defined to be

$$\sum_{n \geq 1} a_n = \lim_{N\to\infty} \sum_{n=1}^{N} a_n$$.​

If the limit exists in $$\mathbb{R}$$ then we say $$\sum \nolimits _{\tiny n \geq 1} a_n$$ is convergent. If the limit does not exist or is $$\pm \infty$$ then $$\sum \nolimits _{\tiny n \geq 1} a_n$$ is called divergent.
Notice that we are not really adding up all the terms in an infinite series at once. We only add up a finite number of the terms and then see how things behave in the limit as the (finite) number of terms tends to infinity: an infinite series is defined to be the limit of its sequence of partial sums.

(Page 3)

For the practical computation of the radius of convergence in basic examples it is convenient to use a new convergence test for positive series.
Theorem 5.5 (Ratio test). If $$a_n \gt 0$$ for all $$n$$, assume

$$q = \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$$​

exists. If $$q \lt 1$$ then $$\sum \nolimits _{\tiny n \geq 1} a_n$$ converges. If $$q \gt 1$$ then $$\sum \nolimits _{\tiny n \geq 1} a_n$$ diverges. If $$q = 1$$ then no conclusion can be drawn.
Proof. Assume $$q \lt 1$$. Pick $$s$$ between $$q$$ and $$1$$: $$q \lt s \lt 1$$. Since $$ a_{n+1}/a_{n} \; \to q$$ ,we have $$a_{n+1}/a_{n} \lt s$$ for all large $$n$$, say for $$n \geq n_0$$. Then $$ a_{n+1} < s a_n$$ when $$n \geq n_0$$, so

$$a_{n_0+1} \lt s a_{n_0} , \; a_{n_0+2} \lt s a_{n_0+1} \lt a^2 a_{n_0} , \; a_{n_0+3} \lt s a_{n_0+2} \lt a^3 a_{n_0} , $$​

and more generally $$a_{n_0+k} < s^k a_{n_0}$$ for any $$k \geq 0$$. Writing $$n_0 + k$$ as $$n$$ we have

$$n \geq n_0 \quad \Rightarrow \quad a_n \lt s^{n-n_0} a_{n_0} = \frac{a_{n_0}}{s^{n_0}} s^n$$.​

Therefore when $$n \geq n_0$$ the number $$a_n$$ is bounded above by a constant multiple of $$s^n$$. Hence $$\sum \nolimits _{\tiny n \geq 1} a_n$$ converges by comparison to a constant multiple of the convergent geometric series $$\sum \nolimits _{\tiny n \geq 1} s^n$$.
In the other direction, if $$q \gt 1$$ then pick $$s$$ with $$1 < s < q$$. An argument similar to the previous one shows $$a_n$$ grows at least as quickly as a constant multiple of $$s^n$$, but this time $$\sum \nolimits _{\tiny n \geq 1} s^n$$ diverges since $$s \gt 1$$. So $$\sum \nolimits _{\tiny n \geq 1} a_n$$ diverges too.
When $$q = 1$$ we can’t make a definite conclusion since both $$\sum \nolimits _{\tiny n \geq 1} 1/n$$ and $$\sum \nolimits _{\tiny n \geq 1} 1/n^2$$ have $$q = 1$$.

(page 16)​

Explicit on page 1, is that understanding limits of infinite sequences of real numbers is a prerequisite -- if you don't understand that topic from analysis then you will make no progress.
Implicit on page 1, is the understanding that there infinite sequences and series are indexed by natural numbers, infinity need not apply.
Explicit on page 3 is a restatement of both of these, including an explicit statement that N only tends to the infinitely large, it doesn't actually get there because infinity is not in the reals and is therefore not in the counting numbers either.
Page 16 leverages both of these understandings to prove the ratio test from understanding of the limit of a geometric series.
Also implicit at the end of the proof on page 1 is that your understanding of limits is defective.

From this we can say that the constant ratio of the infinite geometric series can very well be maintained in the infinity also.

That does not follow as n, N and k never take on values which are infinitely large.

Hence, $$\lim_{n \to \infty} \frac{T_{n+1}}{T_n} \ne \frac{0}{0}$$.

This is a correct statement but only by accident since you don't understand limits of sequences or that n nowhere takes on an infinitely large value and that nothing can "equal" $$\frac{0}{0}$$ since that is not a number.

For all counting numbers n, $$\frac{T_{n+1}}{T_n} = \frac{9 \times 10^{-(n+1)}}{9 \times 10^{-n}} = \frac{1}{10}$$ and thus $$\lim_{n \to \infty} \frac{T_{n+1}}{T_n} = \lim_{n \to \infty} \frac{1}{10} = \frac{1}{10}$$ . That's how this math is done.

Further, Keith Conrad says that you are to assume that both “$$a_n \gt 0$$ for all $$n$$” and “$$q = \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$$ exists [in the real numbers].” So nowhere is $$\frac{0}{0}$$ or $$T_n = 0$$ part of the discussion Keith Conrad is making. Keith Conrad is working with the reals and counting numbers except in the case where infinite series or sequences diverge. Nowhere does Keith Conrad say $$a_{\infty}$$ because that is not sensible notation when $$a_n$$ is indexed by counting numbers. Keith Conrad has to introduce definition 2.4 specifically because there is no way in arithmetic to add up an unending sequence of terms and so he resorts (as in “standard math”) to the language of limits from analysis.

 

[hansda]

For all counting numbers n, $$\frac{T_{n+1}}{T_n} = \frac{9 \times 10^{-(n+1)}}{9 \times 10^{-n}} = \frac{1}{10}$$ and thus $$\lim_{n \to \infty} \frac{T_{n+1}}{T_n} = \lim_{n \to \infty} \frac{1}{10} = \frac{1}{10}$$ . That's how this math is done.

So, $$\lim_{n \to \infty} \frac{T_{n+1}}{T_n} = \frac{1}{10}$$.

Considering the limit theorems [Ref Theorem 400 in association with Remark 401 for item 3. in page 151],
we can also write, $$\lim_{n \to \infty} \frac {T_{n+1}}{T_n} = \frac {\lim_{n \to \infty} T_{n+1}}{\lim_{n \to \infty}T_n} = \frac{T_{\infty +1}}{T_\infty}=\frac{1}{10}$$

As the number $$\frac{1}{10}$$ is not undefined, from the above we can say that $$\lim_{n \to \infty}T_n \ne 0$$.