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

In summary, the conversation is about a specific example from Houshang H. Sohrab's book "Basic Real Analysis" (Second Edition), focusing on Chapter 2: Sequences and Series of Real Numbers. The example discusses the absolute convergence of a series given two sequences. The question is asking for clarification on how/why the series is absolutely convergent. This is proven using an inequality, $(\dagger)$, which shows that the series is less than or equal to the product of the two sequences' norms, which is finite.
  • #1
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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 \(\displaystyle x = (x_n), \ y = (y_n) \in l^2 ( \mathbb{N} )\), the series \(\displaystyle \sum_{ n = 1 }^{ \infty } x_n y_n \) is absolutely convergent ... ..."
My question is as follows:

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

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

Help will be much appreciated ... ...

Peter
 

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  • #2
Peter said:
In the above Example from Sohrab we read the following:

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

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

... does it follow that the series \(\displaystyle \sum_{ n = 1 }^{ \infty } x_n y_n \) is absolutely convergent ... ...?
This comes from the inequality $(\dagger)$, where it is proved that \(\displaystyle \sum_{ n = 1 }^{ \infty }| x_n y_n| \leqslant \|x\|_2\|y\|_2 < \infty. \)
 
  • #3
Opalg said:
This comes from the inequality $(\dagger)$, where it is proved that \(\displaystyle \sum_{ n = 1 }^{ \infty }| x_n y_n| \leqslant \|x\|_2\|y\|_2 < \infty. \)
Appreciate the help, Opalg ...

Thanks ...

Peter
 

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

What is the definition of "l^2(N)"?

"l^2(N)" is a notation used to represent the set of all square-summable real sequences. This means that the sum of the squares of the elements in the sequence must converge to a finite value.

What is the significance of "Sohrab Example 2.3.52" in relation to "l^2(N)"?

"Sohrab Example 2.3.52" is an example given in the book "Real Analysis" by Halsey Royden and Patrick Fitzpatrick. It demonstrates a sequence that belongs to "l^2(N)" but not to "l^1(N)". This highlights the difference between the two sets and the importance of the square-summable property in "l^2(N)".

How is "l^2(N)" related to other sets of real sequences?

"l^2(N)" is a subset of the set of all real sequences, but it is also a superset of "l^1(N)". This means that all elements in "l^2(N)" are also in the set of all real sequences, but not all elements in "l^1(N)" are in "l^2(N)".

Can you provide an example of a sequence that belongs to "l^2(N)"?

One example of a sequence that belongs to "l^2(N)" is the sequence {1, 1/2, 1/3, 1/4, ...}. The sum of the squares of this sequence is 1 + 1/4 + 1/9 + 1/16 + ... = π^2/6, which converges to a finite value.

How is "l^2(N)" used in real-world applications?

"l^2(N)" is commonly used in mathematics and engineering to represent signals or data that have a finite energy or power. It is also used in probability theory to represent square-integrable random variables. In addition, it has applications in Fourier analysis and functional analysis.

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