Convergence of infinite sequences

[Cassi]

Homework Statement

Let V consist of all infinite sequences {x_n} of real numbers for which the series summation x_n² converges. If x = {x_n} and y = {y_n} are two elements of V, define (x,y) = summation (n=1 to infinity) x_ny_n.
Prove that this series converges absolutely.

Homework Equations

The question includes a Hint: Use the Cauchy-Schwarz inequality to estimate the sum, summation (n=1 to M) lx_ny_nl.

The Attempt at a Solution

Using the definition of convergence and ineqaulities I have shown that since x_n converges, we have the inequality, summation(x_n) < summation (x_n²) < infinity. Therefore, {x_n} converges but I do not know how to use the hint to extend this to the inner product.

 

[pasmith]

Homework Statement

Let V consist of all infinite sequences {x_n} of real numbers for which the series summation x_n² converges. If x = {x_n} and y = {y_n} are two elements of V, define (x,y) = summation (n=1 to infinity) x_ny_n.
Prove that this series converges absolutely.

Homework Equations

The question includes a Hint: Use the Cauchy-Schwarz inequality to estimate the sum, summation (n=1 to M) lx_ny_nl.

So what is the Cauchy-Schwartz inequality? Your aim is probably to show that $\sum |x_n y_n| \leq \left(\sum x_n^2\right)^{1/2}\left(\sum y_n^2\right)^{1/2}$. Why is that enough for you to conclude that $\sum |x_ny_n|$ converges?

The Attempt at a Solution

Using the definition of convergence and ineqaulities I have shown that since x_n converges, we have the inequality, summation(x_n) < summation (x_n²) < infinity.

This is false. If $0 < x_n < 1$ then $x_n > x_n^2$; and in particular you should be aware that $\sum_{n=1}^\infty 1/n$ diverges, whereas $\sum_{n=1}^\infty 1/n^2 = \pi^2/6$. Even if it were true it wouldn't assist you.

 

[RUber]

You could also note that for any choice of x and y in V, $|x_n| |y_n| \leq \max( x_n^2, y_n^2 ), \, \forall n \in \mathbb{N}$. This completely disregards your hint, but in my mind is pretty straightforward.