MHB Why is the Series Absolutely Convergent for Sequences in l^2(N)?

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I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 2: Sequences and Series of Real Numbers ... ...

I need help with an aspect of Example 2.3.52 ...

The start of Example 2.3.52 reads as follows ... ...
View attachment 9109
In the above Example from Sohrab we read the following:

" ... ... Then, given any sequences $$x = (x_n), \ y = (y_n) \in l^2 ( \mathbb{N} )$$, the series $$\sum_{ n = 1 }^{ \infty } x_n y_n $$ is absolutely convergent ... ..."
My question is as follows:

How/why, exactly, given any sequences $$x = (x_n), \ y = (y_n) \in l^2 ( \mathbb{N} )$$ ...

... does it follow that the series $$\sum_{ n = 1 }^{ \infty } x_n y_n $$ is absolutely convergent ... ...?

Help will be much appreciated ... ...

Peter
 

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Peter said:
In the above Example from Sohrab we read the following:

" ... ... Then, given any sequences $$x = (x_n), \ y = (y_n) \in l^2 ( \mathbb{N} )$$, the series $$\sum_{ n = 1 }^{ \infty } x_n y_n $$ is absolutely convergent ... ..."
My question is as follows:

How/why, exactly, given any sequences $$x = (x_n), \ y = (y_n) \in l^2 ( \mathbb{N} )$$ ...

... does it follow that the series $$\sum_{ n = 1 }^{ \infty } x_n y_n $$ is absolutely convergent ... ...?
This comes from the inequality $(\dagger)$, where it is proved that $$\sum_{ n = 1 }^{ \infty }| x_n y_n| \leqslant \|x\|_2\|y\|_2 < \infty. $$
 
Opalg said:
This comes from the inequality $(\dagger)$, where it is proved that $$\sum_{ n = 1 }^{ \infty }| x_n y_n| \leqslant \|x\|_2\|y\|_2 < \infty. $$
Appreciate the help, Opalg ...

Thanks ...

Peter
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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