# How can an infinity series be less than infinity?

1. Feb 8, 2009

### Cauchy1789

1. The problem statement, all variables and given/known data

I have a space M which is a sequeces of real numbers $$\{x_n\}$$ where

$$\sum_{n = 1}^{\infty} x_{n}^2 < \infty$$

How can a series mentioned above be become than less than infinity??

Sincerely
Cauchy

2. Feb 8, 2009

### Citan Uzuki

Just because the series has infinitely many terms doesn't mean the sum of all those terms is infinite. For instance, if x_n = 1/2^(n/2), then we have:

$$\sum_{n=1}^{\infty} x_n^2 = \sum_{n=1}^{\infty} \left( \frac{1}{2^{n/2}} \right)^2 = \sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1$$

You might want to do some reading on http://en.wikipedia.org/wiki/Convergent_series" [Broken]

Last edited by a moderator: May 4, 2017
3. Feb 8, 2009

### Cauchy1789

Then my above series can be written as

$$\sum_{n = 1}^{\infty} x_{n}^2 = \lim_{n \to \infty} \int_{1}^{n} x_{n} ^2 dx < \infty$$

Last edited by a moderator: May 4, 2017
4. Feb 8, 2009

### arildno

No.
That is completely nonsensical.

What were you trying to say?

5. Feb 8, 2009

### Cauchy1789

What I would simply like to understand is why a sequence of real numbers defined as in my original post can be less than infinity?

If M is a vectorspace is the "less than infinity" part because a Vectorspace is closed set?

6. Feb 8, 2009

### Citan Uzuki

It doesn't say that the sequence is less than infinity (and indeed, such a statement is nonsensical). It says that the sum of the squares of all the terms in the sequence is less than infinity. And I've already shown that there are sequences satisfying that requirement.

No, the restriction on the sums of the squares is simply part of the definition of the vector space. It has nothing to do with whether the space is closed.

7. Feb 8, 2009

### Cauchy1789

I get it now but anyway.

With the above then if I need to show that M has an inner product defined by

$$\langle{\{x_n\}, \{y_n\}}\rangle = \sum_{n = 1}^{\infty} x_n y_n$$

So what I need to do here isn't to show since the Vector Space M is defined as

$$\sum_{n = 1}^{\infty} x_n^2 < \infty$$ which is the sum of the squared real vector components of the space.

and then if the definition of the inner product applied gives

$$\langle{x_n, x_n}\rangle = \sum_{n = 1}^{\infty} x_n^2$$

then the inner product of two vector sequences of real numbers (if y is defined on M) is $${\{x_n\}, \{y_n\}} = \sum_{n = 1}^{\infty} x_n y_n$$ ???

8. Feb 8, 2009

### Citan Uzuki

You are given a definition, namely that:

$$\langle \vec{x},\ \vec{y} \rangle = \sum_{n=1}^{\infty} x_ny_n$$

where $\vec{x} = \{x_n\}$, and likewise $\vec{y} = \{y_n\}$. What you need to do is show that this function is an inner product on M. Look up the definition of an inner product, and verify that this function satisfies that definition.

9. Feb 8, 2009

### Staff: Mentor

It's not clear to me what you are asked to show. Is it this:
Show that
$$\langle{\{x_n\}, \{y_n\}}\rangle = \sum_{n = 1}^{\infty} x_n y_n$$
is an inner product for M?

If that's the question, you need to verify that the axioms for an inner product hold; namely
symmetry-- $\langle{\{x_n\}, \{y_n\}}\rangle = \langle{\{y_n\}, \{x_n\}}\rangle$
linearity in the first variable--$\langle{\{a*x_n\}, \{y_n\}}\rangle = a\langle{\{x_n\}, \{y_n\}}\rangle$,
and $\langle{\{x_n\}, \{y_n\} + \{z_n\}}\rangle = \langle{\{x_n\}, \{y_n\}}\rangle + \langle{\{x_n\}, \{z_n\}}\rangle$
positive definiteness--$\langle{\{x_n\}, \{x_n\}}\rangle > 0$

10. Feb 8, 2009

### Staff: Mentor

There are a couple of things going on here that you might not be clear on. The space M consists of sequences for which the sum of their squares is finite.

A sequence and its sum are two separate things. A sequence can converge even while the sum of its terms does not.

For example, the sequence {x_n} = {1, 1/2, 1/3, 1/4, ..., 1/n, ...} converges to zero, but the infinite sum of all of its terms diverges (to infinity).

11. Feb 8, 2009

### Cauchy1789

Yes with respect to the definition that M is Vector Space consisting of real numbers with

$$\sum_{n = 1}^{\infty} x_n^2 < \infty$$. This last part that means that the squared normed space on M is less than infinity? Thus the normed space is not infinitely large and then its possible to test the properties of the inner product on it?

12. Feb 8, 2009

### Staff: Mentor

No. This doesn't make any sense. We're not talking about the numeric value of a normed space, or how large geometrically it is, just whether your particular space M has an inner product. You have the definition of <{xn}, {yn}>. Now your job is to show that this inner product satisfies the three axioms I listed. If it does, then your space of infinite sequences is an inner product space.

13. Feb 8, 2009

### Cauchy1789

Question.

Does all axioms of the product have to shown in context with the desired space M in order to show that the inner product and x and y satisfies the definition?

Last edited: Feb 8, 2009
14. Feb 8, 2009

### HallsofIvy

Staff Emeritus
What course is this for? I would think that theorem "if {$x_n$} and {$y_n$} are "square summable" ($\Sigma x_n^2$ and $\Sigma y_n^2$ converge) then $\Sigma x_ny_n$ converges" would be in a fairly advanced course, perhaps "Functional Analysis" but you asking very elementary questions from calculus or precalculus.