Sums over arbitrary index set

  • Thread starter Damidami
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I have read somewhere that we can extend the notion of a series of a sequence
[tex] \sum_{i=1}^{\infty} a_n [/tex]
to sums over an arbitrary index set, say
[tex] a : I \to \mathbb{R} [/tex]
is a family of real number indexed by I, then
[tex] \sum_{i \in I} a_i [/tex]
is the sum of all the elements.

I think the text said that [tex]a_i \geq 0 [/tex] in order to has sense, and that if [tex]a_i \neq 0[/tex] for a non-numerable size of elements, then the series can't converge.

1) My question is with that last sentence, why cant converge such a sequence?, for example if the family is
[tex]a : \mathbb{R_{>0}} \to \mathbb{R}, a(i) = \frac{1}{i}[/tex]
how does one sum over all it's elements?

2) Other question, the fact that all the [tex]a_i \geq 0[/tex] is required because we can't asume an order on the elements of the index set, and if I have negative elements the convergence can vary according to the order in wich I made the sum?

Thanks in advance for any help in understanding this.
 

Answers and Replies

  • #2
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The thing is that we define the convergence slightly different.

If a series [itex]\sum_{n=1}^{+\infty}{a_n}[/itex] has positive terms, then we can say that the series converges if and only if

[tex]\left\{\left.\sum_{k\in K}{a_k}~\right|~K\subseteq \mathbb{N}~\text{finite}\right\}[/tex]

is bounded. (this does not hold for negative terms).

So we mimic this definition. If [itex]I\rightarrow \mathbb{R}:i\rightarrow u_i[/itex] is all positive, then we say that it is "summable" if

[tex]\left\{\left.\sum_{k\in K}{a_k}~\right|~K\subseteq I\right~\text{finite}\}[/tex]

is bounded.

It can now be proven that if [itex]I\rightarrow \mathbb{R}:i\rightarrow u_i[/itex] is "summable", then at most countably many terms are nonzero. Indeed, let's put

[tex]D_n=\{i\in I~\vert~a_i\geq 1/n\}[/tex]

The [itex]D_n[/itex] is finite. Otherwise, we could find a subset of k elements with k>nM with M an upper bound of

[tex]\left\{\left.\sum_{k\in K}{a_k}~\right|~K\subseteq I\right~\text{finite}\}[/tex]

But then

[tex]\sum_{i\in D_n}{a_i}>M[/tex]

which is a contradiction. This implies that the [itex]D_n[/itex] are finite. And thus

[tex]\{i\in I~\vert~a_i\neq 0\}=\bigcup_n{D_n}[/tex]

is countable.

So summable families coincide with convergent series.
 
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  • #3
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Hi micromass,
Thanks, it's pretty clear!
Some questions thought:
1) In all instances when you write
[tex]K \subseteq I[/tex]
do we have to assume that K is finite? (for the sum to have any sense)
2) In the last step when you say
[tex]\displaystyle \cup_n D_n[/tex]
is countable, you are using that a countable union of finite sets is countable?
3) In all cases we are assumming the standard topology in
[tex]R[/tex]
? (the order topology when we say the sum is bounded)? Or isn't a sense of limit needed somehow to say it's summable isn't equivalent to say that it converges?
 
  • #4
22,089
3,294
Hi micromass,
Thanks, it's pretty clear!
Some questions thought:
1) In all instances when you write
[tex]K \subseteq I[/tex]
do we have to assume that K is finite? (for the sum to have any sense)
Yes, I'm sorry. All of it needs to be finite. I forgot that.

2) In the last step when you say
[tex]\displaystyle \cup_n D_n[/tex]
is countable, you are using that a countable union of finite sets is countable?
Yes.

3) In all cases we are assumming the standard topology in
[tex]R[/tex]
? (the order topology when we say the sum is bounded)? Or isn't a sense of limit needed somehow to say it's summable isn't equivalent to say that it converges?
Yes, we assume the standard topology on [itex]\mathbb{R}[/itex].

If you're acquainted with nets, then you'll see that I just described a net here.
 

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