split - Convergence Test Help (unnecessary arguments)

[BigBeachBanana]

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.

From elementary division, 1 divided by 3 is 0.333... or [imath]\frac{1}{3}=0.\bar{3}[/imath]. Do you agree with this?

Moreover, we can represent [imath]0.\bar{3}[/imath] as an infinite series [imath]\sum_{k=1}^{\infty}\frac{3}{10^k}[/imath]

Therefore we say that [imath]\frac{1}{3}[/imath] is exactly equal to [imath]0.\bar{3}[/imath] and can be represented with the infinite series [imath]\sum_{k=1}^{\infty}\frac{3}{10^k}[/imath].

I believe you're arguing the other way around where you say since the limit of the infinite series [imath]\sum_{k=1}^{\infty}\frac{3}{10^k}[/imath] = [imath]0.\bar{3}[/imath], therefore [imath]0.\bar{3}[/imath] is the limit of [imath]\frac{1}{3}[/imath]

Repeating decimals came way before infinitely series even existed (This is what SK was trying to say).
The bar notation is denoting the infinitely repeating decimal places. It does not represent the limit of the infinite series.

If you agree that [imath]\frac{1}{3}=0.\bar{3}[/imath], then multiplying both sides by 3, we get [imath]1=0.\bar{9}.[/imath]

We say 1 is exactly equal to [imath]0.\bar{9}[/imath] and can be represented with the infinite series[imath]\sum_{k=1}^{\infty}\frac{9}{10^k}[/imath].

PS: Chicken came before the egg. ??

 

[Cubist]

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

I follow your logic. And I don't see a particular problem when you state...

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

...except that there are easier ways to evaluate the LHS (as 1)

However your statements in the following chain of posts (above) imply some incorrect thought...

Steven G: ...please tell me any number of your choice that is in between 0.999ˉ and 1 ...
You: I will answer your question if you can tell me how many 9s after the decimal point.
Steven G: There are infinitely many 9s after the decimal point.
You: Infinitely many 9s is not a number. Give me a number. So, how many 9s after the decimal point?

When a sum has infinity written as one (or both) of its end points then there's always an implied limit. This isn't written explicitly because it saves on ink. But, honestly, it IS there

[math] \sum_{k=1}^{\infty}\frac{9}{10^k} \implies \lim_{n \to \infty} \left( \sum_{k=1}^n\frac{9}{10^k} \right) [/math]
You can't just reach "within" this limit to pluck out a finite value of k in order to argue that [imath]\sum_{k=1}^{\infty}\frac{9}{10^k} < 1[/imath]

 

[Dr.Peterson]

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.

What you're missing is that we define the value of a repeating decimal as that limit; so it is correct to say that its value is equal to 1.

Without such a definition, the expression would have no meaning. So the distinction you are making is meaningless.

In exactly the same way, the infinite sum is defined as the limit of partial sums, so we can say that the sum is 1, without needing to explicitly say anything about limits.

 

[nasi112]

I thank a lot everyone who has provided here some of their knowledge. I have no comments on your new posts as I have already explained the idea behind my logic.

But this is not the end of the line. I may still be convinced that, \(\displaystyle 0.\bar{9} = 1\), if Jomo, I mean professor Steven, can tell me how many 9s after the decimal point.

Good luck in your lives.

 

[BigBeachBanana]

I thank a lot everyone who has provided here some of their knowledge. I have no comments on your new posts as I have already explained the idea behind my logic.

But this is not the end of the line. I may still be convinced that, \(\displaystyle 0.\bar{9} = 1\), if Jomo, I mean professor Steven, can tell me how many 9s after the decimal point.

Good luck in your lives.

Maybe Wikipedia can convince you. There are many proofs. Pick one to your liking, including the infinite series one.

 

[topsquark]

I thank a lot everyone who has provided here some of their knowledge. I have no comments on your new posts as I have already explained the idea behind my logic.

But this is not the end of the line. I may still be convinced that, \(\displaystyle 0.\bar{9} = 1\), if Jomo, I mean professor Steven, can tell me how many 9s after the decimal point.

Good luck in your lives.

Sooo.... If there are a finite number of nines after the 0 then it's not equal to 1. But if there are an infinite number of 9's, which is the definition, then it must be wrong because infinity isn't a number. Doesn't it occur to you that you are asking an invalid question?

How many 3's are there after the decimal point in 1/3? An infinite number. There has to be otherwise it won't be equal to 1/3. I presume (I hope!) you agree with this. So how is this any different?

-Dan