split - Convergence Test Help (unnecessary arguments)

[Steven G]

If we do a comparison in infinite series, we don't care about the first few terms. Even without saying eventually, it is understandable.

I understand what you are saying but in my opinion, nothing in math is assumed to be understood, one needs to state any restrictions.

 

[Steven G]

I cannot prove it by myself. This method was provided by our professor and it is guaranteed to always work. Prove me wrong!

1.00001*.99999999999 > 1

Yes, you and your professor are absolutely correct----but only if you have more 0's before than 1 than you have 9's after the decimal.
I think that I already proven you wrong by stating that 1.00001*.99999999999 > 1

 

[AvgStudent]

1.00001*.99999999999 > 1

Since $0.99\bar{9}=1$, then $1.00001\times 0.99\bar{9}=1.00001\times 1 =1.00001>1?$

 

[Steven G]

The idea of this method is that we don't really need to know exactly what happens in $\displaystyle \infty$. It is enough to understand the behavior of the two functions in the way to infinite.

Really? There are functions/series that have a bounded value/sum but at infinity the function/series approaches infinity.
Unfortunately, I can't think of any at the moment. Hopefully someone else will post some functions for us to see.
My mind keeps going to the union and intersection of sets for my example above.

 

[nasi112]

Yes, you and your professor are absolutely correct----but only if you have more 0's before than 1 than you have 9's after the decimal.
I think that I already proven you wrong by stating that 1.00001*.99999999999 > 1

Saying that you have already proven me wrong, means that you have not read my long post.

1.00001*.99999999999 > 1 is one of my professor cases!

Really? There are functions/series that have a bounded value/sum but at infinity the function/series approaches infinity.
Unfortunately, I can't think of any at the moment. Hopefully someone else will post some functions for us to see.
My mind keeps going to the union and intersection of sets for my example above.

Don't forget that this method was used only when,

$\displaystyle \lim_{x \rightarrow \infty} f(x)g(x) = 1$

What you are talking about is another thing.

Since $0.99\bar{9}=1$, then $1.00001\times 0.99\bar{9}=1.00001\times 1 =1.00001>1?$

Believe it or not,
$\displaystyle 0.99\bar{9} \neq 1$

When $\displaystyle 0.99\bar{9}$ is written as $\displaystyle 0.99\bar{9} = 1$, it means that the limit of the sequence of partial sums as we add more and more terms is approaching $\displaystyle 1$.

$\displaystyle 0.99\bar{9} = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000}..... = 9\left(\frac{1}{10}\right) + 9\left(\frac{1}{10^2}\right) + 9\left(\frac{1}{10^3}\right).....=\sum_{k=1}^{\infty}\frac{9}{10^k} = 1$

This is not even a sum in the usual sense of the word, but rather the limit of the sequence of partial sums.

-------
$\displaystyle 1.00001\times 0.99\bar{9} = 1.00001\bar{0}\times 0.99999\bar{9}$

This is case d.

The position of the digit $\displaystyle 1$ is not the last digit as you are comparing,

$\displaystyle 0.99999\bar{9}$
$\displaystyle 1.00001\bar{0}$

$\displaystyle 1.00001\times 0.99\bar{9} > 1$

 

[topsquark]

Believe it or not,
$\displaystyle 0.99\bar{9} \neq 1$

Ummmm..... Yes it does. $\displaystyle \sum_{k= 1}^{\infty} \dfrac{9}{10}$ is perfectly well defined as a limit. Please let's not get into a big argument about this here. There are plenty of other threads on the site that do that.

-Dan

 

[Steven G]

Sorry topsquark, but I just have to ask this question to nasi112.

If you think that 0.999ˉ <1, then please tell me any number of your choice that is in between 0.999ˉ and 1.
If you think that 0.999ˉ >1, then please tell me any number of your choice that is in between 1 and 0.999ˉ.

Unless you can do that, the debate is over.

 

[nasi112]

Sorry topsquark, but I just have to ask this question to nasi112.

If you think that 0.999ˉ <1, then please tell me any number of your choice that is in between 0.999ˉ and 1.
If you think that 0.999ˉ >1, then please tell me any number of your choice that is in between 1 and 0.999ˉ.

Unless you can do that, the debate is over.

I will answer your question if you can tell me how many 9s after the decimal point.

 

[Steven G]

I will answer your question if you can tell me how many 9s after the decimal point.

There are infinitely many 9s after the decimal point.

 

[nasi112]

Infinitely many 9s is not a number. Give me a number. So, how many 9s after the decimal point?

It seems you cannot answer my question!

 

[Steven G]

Infinitely many 9s is not a number. Give me a number. So, how many 9s after the decimal point?

It seems you cannot answer my question!

Believe it or not,
$\displaystyle 0.99\bar{9} \neq 1$

You, not me, wrote that $\displaystyle 0.99\bar{9} \neq 1$
So either $\displaystyle 0.99\bar{9} < 1 or 0.99\bar{9} > 1$
Please pick one and give me a number in between.

 

[topsquark]

Infinitely many 9s is not a number. Give me a number. So, how many 9s after the decimal point?

It seems you cannot answer my question!

The OP is not about this subject! As a suggestion, please refer to this thread.

-Dan

 

[nasi112]

You, not me, wrote that $\displaystyle 0.99\bar{9} \neq 1$
So either $\displaystyle 0.99\bar{9} < 1 or 0.99\bar{9} > 1$
Please pick one and give me a number in between.

Yes. $\displaystyle 0.99\bar{9} \neq 1$.

And $\displaystyle 0.99\bar{9} = 1$ (means the limit of the sequence of the partial sums is approaching $\displaystyle 1$ as you add more and more terms)

Please pick one and give me a number in between.

If you cannot answer my question, how am I supposed to answer your question?

The OP is not about this subject! As a suggestion, please refer to this thread.

-Dan

No need.
I will stop here.


And don't forget that professor Steven G that the main discussion was about $\displaystyle f(x) \times g(x) > 1$ or $\displaystyle f(x) \times g(x) < 1$ when $\displaystyle \lim_{x \rightarrow \infty} f(x)g(x) = 1$, but you jumped outside the box talking about a side post.

 

[Steven G]

Yes. $\displaystyle 0.99\bar{9} \neq 1$.

And $\displaystyle 0.99\bar{9} = 1$ (means the limit of the sequence of the partial sums is approaching $\displaystyle 1$ as you add more and more terms)

A number can not equal two different values.
Since you have not answered my question about finding a number in between I will not respond to any of your posts until you do so.

 

[Cubist]

And $\displaystyle 0.99\bar{9} = 1$ (means the limit of the sequence of the partial sums is approaching $\displaystyle 1$ as you add more and more terms)

I think that a number with a repeating decimal (click) is just a number with a single value as @Steven G says. In particular read converting repeating decimals to fractions on that page (click)

Can you provide any reference to books/ sites that treat repeating/ recurring decimals as limits? (BTW: I'm not being argumentative or condescending, I'm just trying to help you out )

 

[nasi112]

A simple

A number can not equal two different values.
Since you have not answered my question about finding a number in between I will not respond to any of your posts until you do so.

A simple definition that can be accepted worldwide by great mathematicians can force $\displaystyle \neq$ to become $\displaystyle =$.
Not because $\displaystyle \neq$ is wrong, but because they decided that any limit of partial sums will be treated as $\displaystyle =$.

So by definition, it is accepted to write directly,
$\displaystyle 0.99\bar{9} = 1$

In fact,
$\displaystyle 0.99\bar{9} = ***... = 1$

What are these stars and what is going on behind the scene? (They are just the explanation of telling you that: Hey professor Steven, the sums are approaching the limit of $\displaystyle 1$).

I think that a number with a repeating decimal (click) is just a number with a single value as @Steven G says. In particular read converting repeating decimals to fractions on that page (click)

Can you provide any reference to books/ sites that treat repeating/ recurring decimals as limits? (BTW: I'm not being argumentative or condescending, I'm just trying to help you out )

We are not concerned about repeating decimals, but rather about this specific number $\displaystyle 0.99\bar{9}$.

Since $\displaystyle 0.99\bar{9} = \sum_{k=1}^{\infty}\frac{9}{10^k}$.

And $\displaystyle \sum_{k=1}^{\infty}\frac{9}{10^k}$ is the limit of the sequence of partial sums.

Then, $\displaystyle 0.99\bar{9} = 1$, means the sums are approaching the limit of $\displaystyle 1$.

Reference is any Calculus Book.

 

[Cubist]

We are not concerned about repeating decimals, but rather about this specific number $\displaystyle 0.99\bar{9}$.

You wrote your number as a repeating decimal therefore we must surely be concerned with the notation of repeating decimals ?

And $\displaystyle \sum_{k=1}^{\infty}\frac{9}{10^k}$ is the limit of the sequence of partial sums.

Then, $\displaystyle 0.99\bar{9} = 1$, means the sums are approaching the limit of $\displaystyle 1$.

Your "Then" doesn't follow. Surely you intend something along the lines of...

Once the limit is evaluated, $\displaystyle \sum_{k=1}^{\infty}\frac{9}{10^k} = 1 = 1.0 = 2^0 =0.\bar{9}$

As far as I'm aware, the notation $\displaystyle 0.99\bar{9}$ doesn't itself imply a limit. However, there are many limits that when evaluated that do yield a value of 1

Reference is any Calculus Book.

This seems a little glib ?

 

[Deleted member 4993]

10 * $\displaystyle 0.\bar{X}$ - $\displaystyle 0.\bar{X}$ = 9 * $\displaystyle 0.\bar{X}$ .................. and

10 * $\displaystyle 0.\bar{X}$ = $\displaystyle X + 0.\bar{X}$

9 * $\displaystyle 0.\bar{X}$ = $\displaystyle X$

$\displaystyle 0.\bar{X}$ = $\displaystyle X$/9

when $\displaystyle X$ = 9

$\displaystyle 0.\bar{9}$ = 9/9 =1

No need to get tangled up in/with "limits". I was taught this in high school (wayyy before I was introduced to Zeno/Liebniz/Newton).

 

[nasi112]

Hey professors. I respect all of your opinions, and I understand why it is hard for you to follow me.

Even if some of you are not professors, I do respect your knowledge. I do respect even the opinion and knowledge of all students of all levels.

I will explain to you again, and for the last time what is this all about and why $\displaystyle 0.\bar{9} = 1$ is a limit of $\displaystyle 1$, rather than pure $\displaystyle 1$ regardless of the equality sign here.

When I say

$\displaystyle \sum_{k=1}^{3}\frac{9}{10^k} = \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} = 0.999$ (pure sum)

This is called a sum, and I hope that we all agree.

When I say

$\displaystyle \sum_{k=1}^{\infty} a_k = a + a_1 + a2 +.... =$ (finite value)

This is also called a sum, but please focus on this: It is not a sum in the usual sense of the word, but rather, the limit of the sequence of partial sums.

$\displaystyle \sum_{k=1}^{\infty} a_k$ =

This equation says that as we add together more and more terms, the sums are approaching the limit of the smiley face.

 

[Steven G]

What number is between 0.999.... and 1.
That is all I will entertain.

1/3=.333...., multiplying by 3 yields 1=.9999....