Mixing
What is the goal behind defining the notion of summability
$begingroup$
Today we've seen the notion of summability in class.
We say that the sequence $(a_i)_{i in I} $ where $a_i in mathbb{R}$ is summable iff:
$$S((a_i)_{i in I}) = sup { sum_{j in J} a_j mid J subset I, J text{ finite} } in mathbb{R}$$
I know that the notion of summability is there in order to prevent the Riemann rearrangement theorem for series that converge but not absolutely.
From here, it makes sens. Then during the lesson we prove some properties like associativity, linearity, "packet summation"...
And after all of this using the above notion of summable (which is a lot of work for "obvious" facts) we prove the following (assuming there is a bijection $f : mathbb{N} to I$) :
$$S((a_i)_{i in I}) = sum_{i = 0}^infty mid a_{f(i)} mid $$
And here I don't understand at all why we did so much work and effort to defined the notion of summability whereas this is exactly the same as absolutely convergent ?
That's why I don't understand at all now why we defined this definition of summability while it's equivalent to the notion : absolutelty convergent.
Thank you !
real-analysis calculus sequences-and-series summation
asked Jan 17 at 19:03
dghkgfzyukzdghkgfzyukz
16112
$endgroup$
add a comment |
$begingroup$
Today we've seen the notion of summability in class.
We say that the sequence $(a_i)_{i in I} $ where $a_i in mathbb{R}$ is summable iff:
$$S((a_i)_{i in I}) = sup { sum_{j in J} a_j mid J subset I, J text{ finite} } in mathbb{R}$$
I know that the notion of summability is there in order to prevent the Riemann rearrangement theorem for series that converge but not absolutely.
From here, it makes sens. Then during the lesson we prove some properties like associativity, linearity, "packet summation"...
And after all of this using the above notion of summable (which is a lot of work for "obvious" facts) we prove the following (assuming there is a bijection $f : mathbb{N} to I$) :
$$S((a_i)_{i in I}) = sum_{i = 0}^infty mid a_{f(i)} mid $$
And here I don't understand at all why we did so much work and effort to defined the notion of summability whereas this is exactly the same as absolutely convergent ?
That's why I don't understand at all now why we defined this definition of summability while it's equivalent to the notion : absolutelty convergent.
Thank you !
real-analysis calculus sequences-and-series summation
asked Jan 17 at 19:03
dghkgfzyukzdghkgfzyukz
16112
$endgroup$
1
$begingroup$
Suppose that $I$ is the set of nonnegative integers and $a_i = -i$. Then $S((a_i)_{i in I}) = 0$, but $sum_{I=0}^infty | a_i | = infty$. Does this contradict what's written in the question?
$endgroup$
– littleO
Jan 17 at 19:29
$begingroup$
@littleO it’s the sup so we also have : $S((a_i)_{i in I}) = infty$
$endgroup$
– Thinking
Jan 17 at 19:40
1
$begingroup$
@Thinking I think we need absolute value signs around $a_j$ in the definition of a summable sequence.
$endgroup$
– littleO
Jan 17 at 20:14
$begingroup$
@littleO Ok, but with the absolute value signs around $a_j$ we still have $S((a_i)_{i in I}) = infty$, I mean it's the sup not the inf, so it still $infty$ with absolute value or no
$endgroup$
– Thinking
Jan 18 at 17:52
add a comment |
$begingroup$
Today we've seen the notion of summability in class.
We say that the sequence $(a_i)_{i in I} $ where $a_i in mathbb{R}$ is summable iff:
$$S((a_i)_{i in I}) = sup { sum_{j in J} a_j mid J subset I, J text{ finite} } in mathbb{R}$$
I know that the notion of summability is there in order to prevent the Riemann rearrangement theorem for series that converge but not absolutely.
From here, it makes sens. Then during the lesson we prove some properties like associativity, linearity, "packet summation"...
And after all of this using the above notion of summable (which is a lot of work for "obvious" facts) we prove the following (assuming there is a bijection $f : mathbb{N} to I$) :
$$S((a_i)_{i in I}) = sum_{i = 0}^infty mid a_{f(i)} mid $$
And here I don't understand at all why we did so much work and effort to defined the notion of summability whereas this is exactly the same as absolutely convergent ?
That's why I don't understand at all now why we defined this definition of summability while it's equivalent to the notion : absolutelty convergent.
Thank you !
real-analysis calculus sequences-and-series summation
asked Jan 17 at 19:03
dghkgfzyukzdghkgfzyukz
16112
$endgroup$
Today we've seen the notion of summability in class.
We say that the sequence $(a_i)_{i in I} $ where $a_i in mathbb{R}$ is summable iff:
$$S((a_i)_{i in I}) = sup { sum_{j in J} a_j mid J subset I, J text{ finite} } in mathbb{R}$$
I know that the notion of summability is there in order to prevent the Riemann rearrangement theorem for series that converge but not absolutely.
From here, it makes sens. Then during the lesson we prove some properties like associativity, linearity, "packet summation"...
And after all of this using the above notion of summable (which is a lot of work for "obvious" facts) we prove the following (assuming there is a bijection $f : mathbb{N} to I$) :
$$S((a_i)_{i in I}) = sum_{i = 0}^infty mid a_{f(i)} mid $$
And here I don't understand at all why we did so much work and effort to defined the notion of summability whereas this is exactly the same as absolutely convergent ?
That's why I don't understand at all now why we defined this definition of summability while it's equivalent to the notion : absolutelty convergent.
Thank you !
real-analysis calculus sequences-and-series summation
real-analysis calculus sequences-and-series summation
asked Jan 17 at 19:03
dghkgfzyukzdghkgfzyukz
16112
asked Jan 17 at 19:03
dghkgfzyukzdghkgfzyukz
16112
edited Jan 17 at 20:09
dghkgfzyukz
asked Jan 17 at 19:03
dghkgfzyukzdghkgfzyukz
16112
asked Jan 17 at 19:03
dghkgfzyukzdghkgfzyukz
16112
asked Jan 17 at 19:03
dghkgfzyukzdghkgfzyukz
16112
16112
1
$begingroup$
Suppose that $I$ is the set of nonnegative integers and $a_i = -i$. Then $S((a_i)_{i in I}) = 0$, but $sum_{I=0}^infty | a_i | = infty$. Does this contradict what's written in the question?
$endgroup$
– littleO
Jan 17 at 19:29
$begingroup$
@littleO it’s the sup so we also have : $S((a_i)_{i in I}) = infty$
$endgroup$
– Thinking
Jan 17 at 19:40
1
$begingroup$
@Thinking I think we need absolute value signs around $a_j$ in the definition of a summable sequence.
$endgroup$
– littleO
Jan 17 at 20:14
$begingroup$
@littleO Ok, but with the absolute value signs around $a_j$ we still have $S((a_i)_{i in I}) = infty$, I mean it's the sup not the inf, so it still $infty$ with absolute value or no
$endgroup$
– Thinking
Jan 18 at 17:52
add a comment |
1
$begingroup$
Suppose that $I$ is the set of nonnegative integers and $a_i = -i$. Then $S((a_i)_{i in I}) = 0$, but $sum_{I=0}^infty | a_i | = infty$. Does this contradict what's written in the question?
$endgroup$
– littleO
Jan 17 at 19:29
$begingroup$
@littleO it’s the sup so we also have : $S((a_i)_{i in I}) = infty$
$endgroup$
– Thinking
Jan 17 at 19:40
1
$begingroup$
@Thinking I think we need absolute value signs around $a_j$ in the definition of a summable sequence.
$endgroup$
– littleO
Jan 17 at 20:14
$begingroup$
@littleO Ok, but with the absolute value signs around $a_j$ we still have $S((a_i)_{i in I}) = infty$, I mean it's the sup not the inf, so it still $infty$ with absolute value or no
$endgroup$
– Thinking
Jan 18 at 17:52
1
1
$begingroup$
Suppose that $I$ is the set of nonnegative integers and $a_i = -i$. Then $S((a_i)_{i in I}) = 0$, but $sum_{I=0}^infty | a_i | = infty$. Does this contradict what's written in the question?
$endgroup$
– littleO
Jan 17 at 19:29
$begingroup$
Suppose that $I$ is the set of nonnegative integers and $a_i = -i$. Then $S((a_i)_{i in I}) = 0$, but $sum_{I=0}^infty | a_i | = infty$. Does this contradict what's written in the question?
$endgroup$
– littleO
Jan 17 at 19:29
$begingroup$
@littleO it’s the sup so we also have : $S((a_i)_{i in I}) = infty$
$endgroup$
– Thinking
Jan 17 at 19:40
$begingroup$
@littleO it’s the sup so we also have : $S((a_i)_{i in I}) = infty$
$endgroup$
– Thinking
Jan 17 at 19:40
1
1
$begingroup$
@Thinking I think we need absolute value signs around $a_j$ in the definition of a summable sequence.
$endgroup$
– littleO
Jan 17 at 20:14
$begingroup$
@Thinking I think we need absolute value signs around $a_j$ in the definition of a summable sequence.
$endgroup$
– littleO
Jan 17 at 20:14
$begingroup$
@littleO Ok, but with the absolute value signs around $a_j$ we still have $S((a_i)_{i in I}) = infty$, I mean it's the sup not the inf, so it still $infty$ with absolute value or no
$endgroup$
– Thinking
Jan 18 at 17:52
$begingroup$
@littleO Ok, but with the absolute value signs around $a_j$ we still have $S((a_i)_{i in I}) = infty$, I mean it's the sup not the inf, so it still $infty$ with absolute value or no
$endgroup$
– Thinking
Jan 18 at 17:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
What if there isn't a bijection between $I$ and $mathbb{N}$?
This more abstract notion of summability lets us take "uncountable sums." Now, this turns out to not actually be that interesting a notion:
$(*)quad$ Suppose $(a_i)_{iin I}$ is summable. Then ${i: a_inot=0}$ is countable.
That is, a summable sequence is always going to be essentially countable.
But that's a nontrivial fact (and see below for a sketch of a proof). On the face of it, the more abstract definition lets us go beyond the usual notion. And while in this particular setting that's not very interesting, the idea of building up to a very large infinitary operation - that is, not necessarily $mathbb{N}$-indexable - via looking at its finite pieces is useful in a number of contexts.
So I would say that the real value of the more abstract definition above is that it frees us from the seemingly-arbitrary restriction to $mathbb{N}$-indexed sums. While not super useful here (although there can be situations where it simplifies notation in proofs), it's an important conceptual generalization to be capable of.
How do we prove the fact $(*)$ above?
Well, the key point is that $mathbb{R}$ is separable - it has a countable dense subset (namely $mathbb{Q}$). This lets us do some clever "bounding" tricks:
Suppose that $(a_i)_{iin I}$ is summable but ${iin I: a_inot=0}$ is uncountable. WLOG every $a_i$ is positive (why is this WLOG?).
For each positive rational $q$, let $$J_q={iin I: a_i>q}.$$ We have $I=bigcup_{qinmathbb{Q}_{>0}}J_q$; since a countable union of countable sets is countable, we must have $J_q$ be uncountable for some (not unique of course) $qinmathbb{Q}_{>0}$.
But if $Xsubseteq J_q$ is finite, we have $sum_{iin X}a_qge vert Xvertcdot q$; taking larger and larger $X$s, we get unboundedly large values for these sums. So we don't have summability.
answered Jan 17 at 21:42
Noah SchweberNoah Schweber
125k10150288
$endgroup$
$begingroup$
Thank you very much for your answer. It makes more sense now. Do you have an example where we want to sum something but the notion of summability makes it easier to sum this "something" since it's hard to sum this "something" with the restriction to $mathbb{N}$-indexed sum ?
$endgroup$
– dghkgfzyukz
Jan 17 at 22:18
add a comment |
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1 Answer
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$begingroup$
What if there isn't a bijection between $I$ and $mathbb{N}$?
This more abstract notion of summability lets us take "uncountable sums." Now, this turns out to not actually be that interesting a notion:
$(*)quad$ Suppose $(a_i)_{iin I}$ is summable. Then ${i: a_inot=0}$ is countable.
That is, a summable sequence is always going to be essentially countable.
But that's a nontrivial fact (and see below for a sketch of a proof). On the face of it, the more abstract definition lets us go beyond the usual notion. And while in this particular setting that's not very interesting, the idea of building up to a very large infinitary operation - that is, not necessarily $mathbb{N}$-indexable - via looking at its finite pieces is useful in a number of contexts.
So I would say that the real value of the more abstract definition above is that it frees us from the seemingly-arbitrary restriction to $mathbb{N}$-indexed sums. While not super useful here (although there can be situations where it simplifies notation in proofs), it's an important conceptual generalization to be capable of.
How do we prove the fact $(*)$ above?
Well, the key point is that $mathbb{R}$ is separable - it has a countable dense subset (namely $mathbb{Q}$). This lets us do some clever "bounding" tricks:
Suppose that $(a_i)_{iin I}$ is summable but ${iin I: a_inot=0}$ is uncountable. WLOG every $a_i$ is positive (why is this WLOG?).
For each positive rational $q$, let $$J_q={iin I: a_i>q}.$$ We have $I=bigcup_{qinmathbb{Q}_{>0}}J_q$; since a countable union of countable sets is countable, we must have $J_q$ be uncountable for some (not unique of course) $qinmathbb{Q}_{>0}$.
But if $Xsubseteq J_q$ is finite, we have $sum_{iin X}a_qge vert Xvertcdot q$; taking larger and larger $X$s, we get unboundedly large values for these sums. So we don't have summability.
answered Jan 17 at 21:42
Noah SchweberNoah Schweber
125k10150288
$endgroup$
$begingroup$
Thank you very much for your answer. It makes more sense now. Do you have an example where we want to sum something but the notion of summability makes it easier to sum this "something" since it's hard to sum this "something" with the restriction to $mathbb{N}$-indexed sum ?
$endgroup$
– dghkgfzyukz
Jan 17 at 22:18
add a comment |
$begingroup$
What if there isn't a bijection between $I$ and $mathbb{N}$?
This more abstract notion of summability lets us take "uncountable sums." Now, this turns out to not actually be that interesting a notion:
$(*)quad$ Suppose $(a_i)_{iin I}$ is summable. Then ${i: a_inot=0}$ is countable.
That is, a summable sequence is always going to be essentially countable.
But that's a nontrivial fact (and see below for a sketch of a proof). On the face of it, the more abstract definition lets us go beyond the usual notion. And while in this particular setting that's not very interesting, the idea of building up to a very large infinitary operation - that is, not necessarily $mathbb{N}$-indexable - via looking at its finite pieces is useful in a number of contexts.
So I would say that the real value of the more abstract definition above is that it frees us from the seemingly-arbitrary restriction to $mathbb{N}$-indexed sums. While not super useful here (although there can be situations where it simplifies notation in proofs), it's an important conceptual generalization to be capable of.
How do we prove the fact $(*)$ above?
Well, the key point is that $mathbb{R}$ is separable - it has a countable dense subset (namely $mathbb{Q}$). This lets us do some clever "bounding" tricks:
Suppose that $(a_i)_{iin I}$ is summable but ${iin I: a_inot=0}$ is uncountable. WLOG every $a_i$ is positive (why is this WLOG?).
For each positive rational $q$, let $$J_q={iin I: a_i>q}.$$ We have $I=bigcup_{qinmathbb{Q}_{>0}}J_q$; since a countable union of countable sets is countable, we must have $J_q$ be uncountable for some (not unique of course) $qinmathbb{Q}_{>0}$.
But if $Xsubseteq J_q$ is finite, we have $sum_{iin X}a_qge vert Xvertcdot q$; taking larger and larger $X$s, we get unboundedly large values for these sums. So we don't have summability.
answered Jan 17 at 21:42
Noah SchweberNoah Schweber
125k10150288
$endgroup$
$begingroup$
Thank you very much for your answer. It makes more sense now. Do you have an example where we want to sum something but the notion of summability makes it easier to sum this "something" since it's hard to sum this "something" with the restriction to $mathbb{N}$-indexed sum ?
$endgroup$
– dghkgfzyukz
Jan 17 at 22:18
add a comment |
$begingroup$
What if there isn't a bijection between $I$ and $mathbb{N}$?
This more abstract notion of summability lets us take "uncountable sums." Now, this turns out to not actually be that interesting a notion:
$(*)quad$ Suppose $(a_i)_{iin I}$ is summable. Then ${i: a_inot=0}$ is countable.
That is, a summable sequence is always going to be essentially countable.
But that's a nontrivial fact (and see below for a sketch of a proof). On the face of it, the more abstract definition lets us go beyond the usual notion. And while in this particular setting that's not very interesting, the idea of building up to a very large infinitary operation - that is, not necessarily $mathbb{N}$-indexable - via looking at its finite pieces is useful in a number of contexts.
So I would say that the real value of the more abstract definition above is that it frees us from the seemingly-arbitrary restriction to $mathbb{N}$-indexed sums. While not super useful here (although there can be situations where it simplifies notation in proofs), it's an important conceptual generalization to be capable of.
How do we prove the fact $(*)$ above?
Well, the key point is that $mathbb{R}$ is separable - it has a countable dense subset (namely $mathbb{Q}$). This lets us do some clever "bounding" tricks:
Suppose that $(a_i)_{iin I}$ is summable but ${iin I: a_inot=0}$ is uncountable. WLOG every $a_i$ is positive (why is this WLOG?).
For each positive rational $q$, let $$J_q={iin I: a_i>q}.$$ We have $I=bigcup_{qinmathbb{Q}_{>0}}J_q$; since a countable union of countable sets is countable, we must have $J_q$ be uncountable for some (not unique of course) $qinmathbb{Q}_{>0}$.
But if $Xsubseteq J_q$ is finite, we have $sum_{iin X}a_qge vert Xvertcdot q$; taking larger and larger $X$s, we get unboundedly large values for these sums. So we don't have summability.
answered Jan 17 at 21:42
Noah SchweberNoah Schweber
125k10150288
$endgroup$
What if there isn't a bijection between $I$ and $mathbb{N}$?
This more abstract notion of summability lets us take "uncountable sums." Now, this turns out to not actually be that interesting a notion:
$(*)quad$ Suppose $(a_i)_{iin I}$ is summable. Then ${i: a_inot=0}$ is countable.
That is, a summable sequence is always going to be essentially countable.
But that's a nontrivial fact (and see below for a sketch of a proof). On the face of it, the more abstract definition lets us go beyond the usual notion. And while in this particular setting that's not very interesting, the idea of building up to a very large infinitary operation - that is, not necessarily $mathbb{N}$-indexable - via looking at its finite pieces is useful in a number of contexts.
So I would say that the real value of the more abstract definition above is that it frees us from the seemingly-arbitrary restriction to $mathbb{N}$-indexed sums. While not super useful here (although there can be situations where it simplifies notation in proofs), it's an important conceptual generalization to be capable of.
How do we prove the fact $(*)$ above?
Well, the key point is that $mathbb{R}$ is separable - it has a countable dense subset (namely $mathbb{Q}$). This lets us do some clever "bounding" tricks:
Suppose that $(a_i)_{iin I}$ is summable but ${iin I: a_inot=0}$ is uncountable. WLOG every $a_i$ is positive (why is this WLOG?).
For each positive rational $q$, let $$J_q={iin I: a_i>q}.$$ We have $I=bigcup_{qinmathbb{Q}_{>0}}J_q$; since a countable union of countable sets is countable, we must have $J_q$ be uncountable for some (not unique of course) $qinmathbb{Q}_{>0}$.
But if $Xsubseteq J_q$ is finite, we have $sum_{iin X}a_qge vert Xvertcdot q$; taking larger and larger $X$s, we get unboundedly large values for these sums. So we don't have summability.
answered Jan 17 at 21:42
Noah SchweberNoah Schweber
125k10150288
answered Jan 17 at 21:42
Noah SchweberNoah Schweber
125k10150288
answered Jan 17 at 21:42
Noah SchweberNoah Schweber
125k10150288
answered Jan 17 at 21:42
Noah SchweberNoah Schweber
125k10150288
125k10150288
$begingroup$
Thank you very much for your answer. It makes more sense now. Do you have an example where we want to sum something but the notion of summability makes it easier to sum this "something" since it's hard to sum this "something" with the restriction to $mathbb{N}$-indexed sum ?
$endgroup$
– dghkgfzyukz
Jan 17 at 22:18
add a comment |
$begingroup$
Thank you very much for your answer. It makes more sense now. Do you have an example where we want to sum something but the notion of summability makes it easier to sum this "something" since it's hard to sum this "something" with the restriction to $mathbb{N}$-indexed sum ?
$endgroup$
– dghkgfzyukz
Jan 17 at 22:18
$begingroup$
Thank you very much for your answer. It makes more sense now. Do you have an example where we want to sum something but the notion of summability makes it easier to sum this "something" since it's hard to sum this "something" with the restriction to $mathbb{N}$-indexed sum ?
$endgroup$
– dghkgfzyukz
Jan 17 at 22:18
$begingroup$
Thank you very much for your answer. It makes more sense now. Do you have an example where we want to sum something but the notion of summability makes it easier to sum this "something" since it's hard to sum this "something" with the restriction to $mathbb{N}$-indexed sum ?
$endgroup$
– dghkgfzyukz
Jan 17 at 22:18
add a comment |
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$begingroup$
Suppose that $I$ is the set of nonnegative integers and $a_i = -i$. Then $S((a_i)_{i in I}) = 0$, but $sum_{I=0}^infty | a_i | = infty$. Does this contradict what's written in the question?
$endgroup$
– littleO
Jan 17 at 19:29
$begingroup$
@littleO it’s the sup so we also have : $S((a_i)_{i in I}) = infty$
$endgroup$
– Thinking
Jan 17 at 19:40
1
$begingroup$
@Thinking I think we need absolute value signs around $a_j$ in the definition of a summable sequence.
$endgroup$
– littleO
Jan 17 at 20:14
$begingroup$
@littleO Ok, but with the absolute value signs around $a_j$ we still have $S((a_i)_{i in I}) = infty$, I mean it's the sup not the inf, so it still $infty$ with absolute value or no
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– Thinking
Jan 18 at 17:52