Why Does the Summation Converge in l_1(R) but Not for the Sequence (1)?

[bugatti79]

Im trying to understand the following. We have l_1(R)=( x=x_n in l(R): summation from n=1 to infinity for absolute value of x_n). It says that this summation converges, but converges to what?

Also , its says (1) is not in l_1(R) but 1/n^2 is. Can some one explain how these are so.

This is not homework, just trying 2 gain an understanding. Thanks.

 

[mathman]

Please clarify. What is l_1(R)? What is (1)? Also your parenthetical expression ( x=x_n etc.) is not very clear.

 

[bugatti79]

Please clarify. What is l_1(R)? What is (1)? Also your parenthetical expression ( x=x_n etc.) is not very clear.

Ok, I will include a few additional lines previous to it. We are examining norms on sequences spaces.
The notes say that $|| ||_1$ cannot be extended to the set of all real sequences
ie $x=(1)=(1,1,1...)$ $||x||_1=1+1+1+1$ does not exist. What does this mean?

So we define

$l_1(R)=\{x=(x_n) \in l(R) : Ʃ |x_n|\}$ from n=1 to infinity.
It says this summation converges, but to what?

Then it says
eg $(1) \notin l_1(R)$ and $1/n^2 \in l_1(R)$
How are these determined?

Thanks

 

[Mark44]

Ok, I will include a few additional lines previous to it. We are examining norms on sequences spaces.
The notes say that $|| ||_1$ cannot be extended to the set of all real sequences
ie $x=(1)=(1,1,1...)$ $||x||_1=1+1+1+1$ does not exist. What does this mean?

You're being a little sloppy here.
As you have defined x, ||x||₁ = 1 + 1 + 1 + ... + 1 + ..., not 1 + 1 + 1 + 1, which equals 4.

The expression "does not exist" here means that ||x||₁ is not a finite number. I.e., the sum is not finite.

So we define

$l_1(R)=\{x=(x_n) \in l(R) : Ʃ |x_n|\}$ from n=1 to infinity.
It says this summation converges, but to what?

Have you omitted something here? The usual way of saying a summation converges is to write it as Ʃ |x_n| < ∞.

Then it says
eg $(1) \notin l_1(R)$ and $1/n^2 \in l_1(R)$
How are these determined?

Ʃ 1 = 1 + 1 + 1 + ..., which is divergent.
Ʃ 1/n² is a well known convergent series (a p-series, with p = 2).

 

[bugatti79]

You're being a little sloppy here.
As you have defined x, ||x||₁ = 1 + 1 + 1 + ... + 1 + ..., not 1 + 1 + 1 + 1, which equals 4.

Ok

The expression "does not exist" here means that ||x||₁ is not a finite number. I.e., the sum is not finite.
Have you omitted something here? The usual way of saying a summation converges is to write it as Ʃ |x_n| < ∞.

I am looking at some ones notes. Yes you are correct. It appears in another similar example.
It seems against my intuition. Isn't it not approaching infinity? How would it converge to < infinity?

Ʃ 1 = 1 + 1 + 1 + ..., which is divergent.
Ʃ 1/n² is a well known convergent series (a p-series, with p = 2).

 

[Mark44]

I am looking at some ones notes. Yes you are correct. It appears in another similar example.
It seems against my intuition. Isn't it not approaching infinity? How would it converge to < infinity?

I gave two examples, so I'm not sure which one you're referring to by "it."
Ʃ1 diverges
Ʃ1/n² converges, so Ʃ1/n² < ∞.

 

[bugatti79]

I gave two examples, so I'm not sure which one you're referring to by "it."
Ʃ1 diverges
Ʃ1/n² converges, so Ʃ1/n² < ∞.

Ok, I think I get you now. Thanks