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Assume I'm unhappy with the Riemann Rearrangement Theorem. How else can I define a series?
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I have been explaining the Riemann Rearrangement Theorem to a friend of mine, and they feel as though the definition of series using partial sums "doesn't work" for conditionally convergent sequences. I understand how they feel: the RRT feels counter-intuitive, and so the result should be denied as a contradiction and the partial-sum definition rejected.
However, the definition of a series as
$$sum_k=1^infty a_k = lim_ntoinfty left(sum_k=1^n a_kright)$$
feels like the most natural approach to take. If I wanted to add infinitely many numbers together by hand, this is how I would have to do it.
Are there an alternative (but not equivalent) definition of an infinite sum that agrees with the partial-sum definition on absolutely convergent sequences, but where the RRT doesn't hold?
My guess is that if you define some value for an infinite series agreeing on absolutely convergent sequences then this method of definition must imply the RRT, but I don't see how that proof could go.
real-analysis sequences-and-series convergence
asked 3 hours ago
Santana Afton
2,3472527
add a comment |Â
up vote
3
down vote
favorite
I have been explaining the Riemann Rearrangement Theorem to a friend of mine, and they feel as though the definition of series using partial sums "doesn't work" for conditionally convergent sequences. I understand how they feel: the RRT feels counter-intuitive, and so the result should be denied as a contradiction and the partial-sum definition rejected.
However, the definition of a series as
$$sum_k=1^infty a_k = lim_ntoinfty left(sum_k=1^n a_kright)$$
feels like the most natural approach to take. If I wanted to add infinitely many numbers together by hand, this is how I would have to do it.
Are there an alternative (but not equivalent) definition of an infinite sum that agrees with the partial-sum definition on absolutely convergent sequences, but where the RRT doesn't hold?
My guess is that if you define some value for an infinite series agreeing on absolutely convergent sequences then this method of definition must imply the RRT, but I don't see how that proof could go.
real-analysis sequences-and-series convergence
asked 3 hours ago
Santana Afton
2,3472527
Since $mathcalP(mathbbN)$ and $mathcalP_fin(mathbbN)$ have different cardinalities ($2^aleph_0$ and $aleph_0$), I believe there is no way to avoid the RRT-lack-of-commutativity phenomenon when defining series (i.e. infinite sums) through finite sums, no matter if in the standard way, through smoothed/Cesà ro sums or whatever.
– Jack D'Aurizio
3 hours ago
I think you're pretty much stuck. There are a huge number of really general notions of summability (e.g. matrix summability methods), and for quite a few of these not only do you still run into these problems, but in fact for pretty much any of these summability methods there are many series for which most rearrangements (in the Baire category sense for certain metrics) of that series have very bad convergence properties (such as every extended real number being a subsequence limit). Or something to this effect . . .
– Dave L. Renfro
3 hours ago
You need to explain to your friend sometimes things make us unhappy and there is not much to do about them.
– Pedro Tamaroff♦
2 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have been explaining the Riemann Rearrangement Theorem to a friend of mine, and they feel as though the definition of series using partial sums "doesn't work" for conditionally convergent sequences. I understand how they feel: the RRT feels counter-intuitive, and so the result should be denied as a contradiction and the partial-sum definition rejected.
However, the definition of a series as
$$sum_k=1^infty a_k = lim_ntoinfty left(sum_k=1^n a_kright)$$
feels like the most natural approach to take. If I wanted to add infinitely many numbers together by hand, this is how I would have to do it.
Are there an alternative (but not equivalent) definition of an infinite sum that agrees with the partial-sum definition on absolutely convergent sequences, but where the RRT doesn't hold?
My guess is that if you define some value for an infinite series agreeing on absolutely convergent sequences then this method of definition must imply the RRT, but I don't see how that proof could go.
real-analysis sequences-and-series convergence
asked 3 hours ago
Santana Afton
2,3472527
I have been explaining the Riemann Rearrangement Theorem to a friend of mine, and they feel as though the definition of series using partial sums "doesn't work" for conditionally convergent sequences. I understand how they feel: the RRT feels counter-intuitive, and so the result should be denied as a contradiction and the partial-sum definition rejected.
However, the definition of a series as
$$sum_k=1^infty a_k = lim_ntoinfty left(sum_k=1^n a_kright)$$
feels like the most natural approach to take. If I wanted to add infinitely many numbers together by hand, this is how I would have to do it.
Are there an alternative (but not equivalent) definition of an infinite sum that agrees with the partial-sum definition on absolutely convergent sequences, but where the RRT doesn't hold?
My guess is that if you define some value for an infinite series agreeing on absolutely convergent sequences then this method of definition must imply the RRT, but I don't see how that proof could go.
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
asked 3 hours ago
Santana Afton
2,3472527
asked 3 hours ago
Santana Afton
2,3472527
asked 3 hours ago
Santana Afton
2,3472527
asked 3 hours ago
Santana Afton
2,3472527
asked 3 hours ago
Santana Afton
2,3472527
2,3472527
Since $mathcalP(mathbbN)$ and $mathcalP_fin(mathbbN)$ have different cardinalities ($2^aleph_0$ and $aleph_0$), I believe there is no way to avoid the RRT-lack-of-commutativity phenomenon when defining series (i.e. infinite sums) through finite sums, no matter if in the standard way, through smoothed/Cesà ro sums or whatever.
– Jack D'Aurizio
3 hours ago
I think you're pretty much stuck. There are a huge number of really general notions of summability (e.g. matrix summability methods), and for quite a few of these not only do you still run into these problems, but in fact for pretty much any of these summability methods there are many series for which most rearrangements (in the Baire category sense for certain metrics) of that series have very bad convergence properties (such as every extended real number being a subsequence limit). Or something to this effect . . .
– Dave L. Renfro
3 hours ago
You need to explain to your friend sometimes things make us unhappy and there is not much to do about them.
– Pedro Tamaroff♦
2 hours ago
add a comment |Â
Since $mathcalP(mathbbN)$ and $mathcalP_fin(mathbbN)$ have different cardinalities ($2^aleph_0$ and $aleph_0$), I believe there is no way to avoid the RRT-lack-of-commutativity phenomenon when defining series (i.e. infinite sums) through finite sums, no matter if in the standard way, through smoothed/Cesà ro sums or whatever.
– Jack D'Aurizio
3 hours ago
I think you're pretty much stuck. There are a huge number of really general notions of summability (e.g. matrix summability methods), and for quite a few of these not only do you still run into these problems, but in fact for pretty much any of these summability methods there are many series for which most rearrangements (in the Baire category sense for certain metrics) of that series have very bad convergence properties (such as every extended real number being a subsequence limit). Or something to this effect . . .
– Dave L. Renfro
3 hours ago
You need to explain to your friend sometimes things make us unhappy and there is not much to do about them.
– Pedro Tamaroff♦
2 hours ago
Since $mathcalP(mathbbN)$ and $mathcalP_fin(mathbbN)$ have different cardinalities ($2^aleph_0$ and $aleph_0$), I believe there is no way to avoid the RRT-lack-of-commutativity phenomenon when defining series (i.e. infinite sums) through finite sums, no matter if in the standard way, through smoothed/Cesà ro sums or whatever.
– Jack D'Aurizio
3 hours ago
Since $mathcalP(mathbbN)$ and $mathcalP_fin(mathbbN)$ have different cardinalities ($2^aleph_0$ and $aleph_0$), I believe there is no way to avoid the RRT-lack-of-commutativity phenomenon when defining series (i.e. infinite sums) through finite sums, no matter if in the standard way, through smoothed/Cesà ro sums or whatever.
– Jack D'Aurizio
3 hours ago
I think you're pretty much stuck. There are a huge number of really general notions of summability (e.g. matrix summability methods), and for quite a few of these not only do you still run into these problems, but in fact for pretty much any of these summability methods there are many series for which most rearrangements (in the Baire category sense for certain metrics) of that series have very bad convergence properties (such as every extended real number being a subsequence limit). Or something to this effect . . .
– Dave L. Renfro
3 hours ago
I think you're pretty much stuck. There are a huge number of really general notions of summability (e.g. matrix summability methods), and for quite a few of these not only do you still run into these problems, but in fact for pretty much any of these summability methods there are many series for which most rearrangements (in the Baire category sense for certain metrics) of that series have very bad convergence properties (such as every extended real number being a subsequence limit). Or something to this effect . . .
– Dave L. Renfro
3 hours ago
You need to explain to your friend sometimes things make us unhappy and there is not much to do about them.
– Pedro Tamaroff♦
2 hours ago
You need to explain to your friend sometimes things make us unhappy and there is not much to do about them.
– Pedro Tamaroff♦
2 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
You could define $L$ to be the infimum of the set of all sums of finitely many terms of the series (not necessarily consecutive terms); and define $U$ to be the supremum of the same set. Then a series would be convergent if both $L$ and $U$ were finite and the sum would be $U-L$.
This would amount to defining a series to be convergent if and only if its positive-term subseries and negative-term subseries each converged (in the traditional sense) to a finite value. ($L$ would be the old-fashioned sum of the negative terms; $U$ of the positive.)
Edit: Correction: The sum should be $U+L$. (E.g. If the positive terms add to $4$ and the negative terms add to $-7$, the sum would be $-3$.)
answered 3 hours ago
paw88789
28.6k12348
Interesting. Can you show this is different from absolute convergence (if it is)?
– Ethan Bolker
3 hours ago
3
I'm pretty sure it isn't different than absolute convergence.
– paw88789
3 hours ago
This doesn't work at all, even for the nicely convergent sequence $a_n=2^-n$. You have $L=0$ and $U=2$ so it doesn't converge. For any conditionally convergent series you will have $L=-infty, U=infty$
– Ross Millikan
2 hours ago
@RossMillikan So for $2^-n$ (for $n=0,1,2,...$ as you say, $U=2$ and $L=0$ so the sum would be $2$. I did have a mistake though, the sum should be $U+L$ not $U-L$, which I have noted in an edit.
– paw88789
2 hours ago
I was responding when you wanted $U=L$ and the sum to be that value. The new version is equivalent to absolute convergence, because the sum of the absolute values of the terms is $U-L$ and if that is finite you have absolute convergence.
– Ross Millikan
17 mins ago
add a comment |Â
up vote
1
down vote
Assume that we want a notion of summability $sum'$ such that
- (Compatibility with absolutely convergent series)
$$sum_ngeq 1|a_n|<+inftyquadLongrightarrowquadsum_ngeq 1'|a_n|=sum_ngeq 1|a_n|$$ - (Compatibility with the vector space operations) Provided that both $sum'a_n$ and $sum' b_n$ are finite, $sum'(a_n+lambda b_n)=sum' a_n+lambdasum' b_n$
- (Negation of Riemann-Dini) Provided that $sum_ngeq 0'a_n$ is finite, for any bijective $sigma:mathbbNtomathbbN$
$$ sum_ngeq 0'a_n = sum_ngeq 0'a_sigma(n)$$
Then such notion of summability is precisely the notion of absolute-summability. Assume that $sum'a_n$ is finite and $a_n$ has an infinite number of both positive and negative terms. By $3.$ we may assume without loss of generality that the sign of $a_n$ agrees with the parity of $n$. If $sum a_n$ is not absolutely convergent then $left|sum a_2nright|$ or $left|sum a_2n+1right|$ is unbounded (or both are). By 2. and 3. both $sum' a_2n$ and $sum' a_2n+1$ have to be finite. Assuming that $|sum a_2n|$ is unbounded, by 2. and 3. again
$$ sum_ngeq 1'|a_2n|geq sum_n=1^N|a_2n| $$
has to hold for any $NinmathbbN^+$, but that implies $left|sum' a_2nright|=+infty.$
answered 3 hours ago
Jack D'Aurizio
279k33272648
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
You could define $L$ to be the infimum of the set of all sums of finitely many terms of the series (not necessarily consecutive terms); and define $U$ to be the supremum of the same set. Then a series would be convergent if both $L$ and $U$ were finite and the sum would be $U-L$.
This would amount to defining a series to be convergent if and only if its positive-term subseries and negative-term subseries each converged (in the traditional sense) to a finite value. ($L$ would be the old-fashioned sum of the negative terms; $U$ of the positive.)
Edit: Correction: The sum should be $U+L$. (E.g. If the positive terms add to $4$ and the negative terms add to $-7$, the sum would be $-3$.)
answered 3 hours ago
paw88789
28.6k12348
Interesting. Can you show this is different from absolute convergence (if it is)?
– Ethan Bolker
3 hours ago
3
I'm pretty sure it isn't different than absolute convergence.
– paw88789
3 hours ago
This doesn't work at all, even for the nicely convergent sequence $a_n=2^-n$. You have $L=0$ and $U=2$ so it doesn't converge. For any conditionally convergent series you will have $L=-infty, U=infty$
– Ross Millikan
2 hours ago
@RossMillikan So for $2^-n$ (for $n=0,1,2,...$ as you say, $U=2$ and $L=0$ so the sum would be $2$. I did have a mistake though, the sum should be $U+L$ not $U-L$, which I have noted in an edit.
– paw88789
2 hours ago
I was responding when you wanted $U=L$ and the sum to be that value. The new version is equivalent to absolute convergence, because the sum of the absolute values of the terms is $U-L$ and if that is finite you have absolute convergence.
– Ross Millikan
17 mins ago
add a comment |Â
up vote
3
down vote
You could define $L$ to be the infimum of the set of all sums of finitely many terms of the series (not necessarily consecutive terms); and define $U$ to be the supremum of the same set. Then a series would be convergent if both $L$ and $U$ were finite and the sum would be $U-L$.
This would amount to defining a series to be convergent if and only if its positive-term subseries and negative-term subseries each converged (in the traditional sense) to a finite value. ($L$ would be the old-fashioned sum of the negative terms; $U$ of the positive.)
Edit: Correction: The sum should be $U+L$. (E.g. If the positive terms add to $4$ and the negative terms add to $-7$, the sum would be $-3$.)
answered 3 hours ago
paw88789
28.6k12348
Interesting. Can you show this is different from absolute convergence (if it is)?
– Ethan Bolker
3 hours ago
3
I'm pretty sure it isn't different than absolute convergence.
– paw88789
3 hours ago
This doesn't work at all, even for the nicely convergent sequence $a_n=2^-n$. You have $L=0$ and $U=2$ so it doesn't converge. For any conditionally convergent series you will have $L=-infty, U=infty$
– Ross Millikan
2 hours ago
@RossMillikan So for $2^-n$ (for $n=0,1,2,...$ as you say, $U=2$ and $L=0$ so the sum would be $2$. I did have a mistake though, the sum should be $U+L$ not $U-L$, which I have noted in an edit.
– paw88789
2 hours ago
I was responding when you wanted $U=L$ and the sum to be that value. The new version is equivalent to absolute convergence, because the sum of the absolute values of the terms is $U-L$ and if that is finite you have absolute convergence.
– Ross Millikan
17 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You could define $L$ to be the infimum of the set of all sums of finitely many terms of the series (not necessarily consecutive terms); and define $U$ to be the supremum of the same set. Then a series would be convergent if both $L$ and $U$ were finite and the sum would be $U-L$.
This would amount to defining a series to be convergent if and only if its positive-term subseries and negative-term subseries each converged (in the traditional sense) to a finite value. ($L$ would be the old-fashioned sum of the negative terms; $U$ of the positive.)
Edit: Correction: The sum should be $U+L$. (E.g. If the positive terms add to $4$ and the negative terms add to $-7$, the sum would be $-3$.)
answered 3 hours ago
paw88789
28.6k12348
You could define $L$ to be the infimum of the set of all sums of finitely many terms of the series (not necessarily consecutive terms); and define $U$ to be the supremum of the same set. Then a series would be convergent if both $L$ and $U$ were finite and the sum would be $U-L$.
This would amount to defining a series to be convergent if and only if its positive-term subseries and negative-term subseries each converged (in the traditional sense) to a finite value. ($L$ would be the old-fashioned sum of the negative terms; $U$ of the positive.)
Edit: Correction: The sum should be $U+L$. (E.g. If the positive terms add to $4$ and the negative terms add to $-7$, the sum would be $-3$.)
answered 3 hours ago
paw88789
28.6k12348
edited 2 hours ago
answered 3 hours ago
paw88789
28.6k12348
answered 3 hours ago
paw88789
28.6k12348
answered 3 hours ago
paw88789
28.6k12348
28.6k12348
Interesting. Can you show this is different from absolute convergence (if it is)?
– Ethan Bolker
3 hours ago
3
I'm pretty sure it isn't different than absolute convergence.
– paw88789
3 hours ago
This doesn't work at all, even for the nicely convergent sequence $a_n=2^-n$. You have $L=0$ and $U=2$ so it doesn't converge. For any conditionally convergent series you will have $L=-infty, U=infty$
– Ross Millikan
2 hours ago
@RossMillikan So for $2^-n$ (for $n=0,1,2,...$ as you say, $U=2$ and $L=0$ so the sum would be $2$. I did have a mistake though, the sum should be $U+L$ not $U-L$, which I have noted in an edit.
– paw88789
2 hours ago
I was responding when you wanted $U=L$ and the sum to be that value. The new version is equivalent to absolute convergence, because the sum of the absolute values of the terms is $U-L$ and if that is finite you have absolute convergence.
– Ross Millikan
17 mins ago
add a comment |Â
Interesting. Can you show this is different from absolute convergence (if it is)?
– Ethan Bolker
3 hours ago
3
I'm pretty sure it isn't different than absolute convergence.
– paw88789
3 hours ago
This doesn't work at all, even for the nicely convergent sequence $a_n=2^-n$. You have $L=0$ and $U=2$ so it doesn't converge. For any conditionally convergent series you will have $L=-infty, U=infty$
– Ross Millikan
2 hours ago
@RossMillikan So for $2^-n$ (for $n=0,1,2,...$ as you say, $U=2$ and $L=0$ so the sum would be $2$. I did have a mistake though, the sum should be $U+L$ not $U-L$, which I have noted in an edit.
– paw88789
2 hours ago
I was responding when you wanted $U=L$ and the sum to be that value. The new version is equivalent to absolute convergence, because the sum of the absolute values of the terms is $U-L$ and if that is finite you have absolute convergence.
– Ross Millikan
17 mins ago
Interesting. Can you show this is different from absolute convergence (if it is)?
– Ethan Bolker
3 hours ago
Interesting. Can you show this is different from absolute convergence (if it is)?
– Ethan Bolker
3 hours ago
3
3
I'm pretty sure it isn't different than absolute convergence.
– paw88789
3 hours ago
I'm pretty sure it isn't different than absolute convergence.
– paw88789
3 hours ago
This doesn't work at all, even for the nicely convergent sequence $a_n=2^-n$. You have $L=0$ and $U=2$ so it doesn't converge. For any conditionally convergent series you will have $L=-infty, U=infty$
– Ross Millikan
2 hours ago
This doesn't work at all, even for the nicely convergent sequence $a_n=2^-n$. You have $L=0$ and $U=2$ so it doesn't converge. For any conditionally convergent series you will have $L=-infty, U=infty$
– Ross Millikan
2 hours ago
@RossMillikan So for $2^-n$ (for $n=0,1,2,...$ as you say, $U=2$ and $L=0$ so the sum would be $2$. I did have a mistake though, the sum should be $U+L$ not $U-L$, which I have noted in an edit.
– paw88789
2 hours ago
@RossMillikan So for $2^-n$ (for $n=0,1,2,...$ as you say, $U=2$ and $L=0$ so the sum would be $2$. I did have a mistake though, the sum should be $U+L$ not $U-L$, which I have noted in an edit.
– paw88789
2 hours ago
I was responding when you wanted $U=L$ and the sum to be that value. The new version is equivalent to absolute convergence, because the sum of the absolute values of the terms is $U-L$ and if that is finite you have absolute convergence.
– Ross Millikan
17 mins ago
I was responding when you wanted $U=L$ and the sum to be that value. The new version is equivalent to absolute convergence, because the sum of the absolute values of the terms is $U-L$ and if that is finite you have absolute convergence.
– Ross Millikan
17 mins ago
add a comment |Â
up vote
1
down vote
Assume that we want a notion of summability $sum'$ such that
- (Compatibility with absolutely convergent series)
$$sum_ngeq 1|a_n|<+inftyquadLongrightarrowquadsum_ngeq 1'|a_n|=sum_ngeq 1|a_n|$$ - (Compatibility with the vector space operations) Provided that both $sum'a_n$ and $sum' b_n$ are finite, $sum'(a_n+lambda b_n)=sum' a_n+lambdasum' b_n$
- (Negation of Riemann-Dini) Provided that $sum_ngeq 0'a_n$ is finite, for any bijective $sigma:mathbbNtomathbbN$
$$ sum_ngeq 0'a_n = sum_ngeq 0'a_sigma(n)$$
Then such notion of summability is precisely the notion of absolute-summability. Assume that $sum'a_n$ is finite and $a_n$ has an infinite number of both positive and negative terms. By $3.$ we may assume without loss of generality that the sign of $a_n$ agrees with the parity of $n$. If $sum a_n$ is not absolutely convergent then $left|sum a_2nright|$ or $left|sum a_2n+1right|$ is unbounded (or both are). By 2. and 3. both $sum' a_2n$ and $sum' a_2n+1$ have to be finite. Assuming that $|sum a_2n|$ is unbounded, by 2. and 3. again
$$ sum_ngeq 1'|a_2n|geq sum_n=1^N|a_2n| $$
has to hold for any $NinmathbbN^+$, but that implies $left|sum' a_2nright|=+infty.$
answered 3 hours ago
Jack D'Aurizio
279k33272648
add a comment |Â
up vote
1
down vote
Assume that we want a notion of summability $sum'$ such that
- (Compatibility with absolutely convergent series)
$$sum_ngeq 1|a_n|<+inftyquadLongrightarrowquadsum_ngeq 1'|a_n|=sum_ngeq 1|a_n|$$ - (Compatibility with the vector space operations) Provided that both $sum'a_n$ and $sum' b_n$ are finite, $sum'(a_n+lambda b_n)=sum' a_n+lambdasum' b_n$
- (Negation of Riemann-Dini) Provided that $sum_ngeq 0'a_n$ is finite, for any bijective $sigma:mathbbNtomathbbN$
$$ sum_ngeq 0'a_n = sum_ngeq 0'a_sigma(n)$$
Then such notion of summability is precisely the notion of absolute-summability. Assume that $sum'a_n$ is finite and $a_n$ has an infinite number of both positive and negative terms. By $3.$ we may assume without loss of generality that the sign of $a_n$ agrees with the parity of $n$. If $sum a_n$ is not absolutely convergent then $left|sum a_2nright|$ or $left|sum a_2n+1right|$ is unbounded (or both are). By 2. and 3. both $sum' a_2n$ and $sum' a_2n+1$ have to be finite. Assuming that $|sum a_2n|$ is unbounded, by 2. and 3. again
$$ sum_ngeq 1'|a_2n|geq sum_n=1^N|a_2n| $$
has to hold for any $NinmathbbN^+$, but that implies $left|sum' a_2nright|=+infty.$
answered 3 hours ago
Jack D'Aurizio
279k33272648
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Assume that we want a notion of summability $sum'$ such that
- (Compatibility with absolutely convergent series)
$$sum_ngeq 1|a_n|<+inftyquadLongrightarrowquadsum_ngeq 1'|a_n|=sum_ngeq 1|a_n|$$ - (Compatibility with the vector space operations) Provided that both $sum'a_n$ and $sum' b_n$ are finite, $sum'(a_n+lambda b_n)=sum' a_n+lambdasum' b_n$
- (Negation of Riemann-Dini) Provided that $sum_ngeq 0'a_n$ is finite, for any bijective $sigma:mathbbNtomathbbN$
$$ sum_ngeq 0'a_n = sum_ngeq 0'a_sigma(n)$$
Then such notion of summability is precisely the notion of absolute-summability. Assume that $sum'a_n$ is finite and $a_n$ has an infinite number of both positive and negative terms. By $3.$ we may assume without loss of generality that the sign of $a_n$ agrees with the parity of $n$. If $sum a_n$ is not absolutely convergent then $left|sum a_2nright|$ or $left|sum a_2n+1right|$ is unbounded (or both are). By 2. and 3. both $sum' a_2n$ and $sum' a_2n+1$ have to be finite. Assuming that $|sum a_2n|$ is unbounded, by 2. and 3. again
$$ sum_ngeq 1'|a_2n|geq sum_n=1^N|a_2n| $$
has to hold for any $NinmathbbN^+$, but that implies $left|sum' a_2nright|=+infty.$
answered 3 hours ago
Jack D'Aurizio
279k33272648
Assume that we want a notion of summability $sum'$ such that
- (Compatibility with absolutely convergent series)
$$sum_ngeq 1|a_n|<+inftyquadLongrightarrowquadsum_ngeq 1'|a_n|=sum_ngeq 1|a_n|$$ - (Compatibility with the vector space operations) Provided that both $sum'a_n$ and $sum' b_n$ are finite, $sum'(a_n+lambda b_n)=sum' a_n+lambdasum' b_n$
- (Negation of Riemann-Dini) Provided that $sum_ngeq 0'a_n$ is finite, for any bijective $sigma:mathbbNtomathbbN$
$$ sum_ngeq 0'a_n = sum_ngeq 0'a_sigma(n)$$
Then such notion of summability is precisely the notion of absolute-summability. Assume that $sum'a_n$ is finite and $a_n$ has an infinite number of both positive and negative terms. By $3.$ we may assume without loss of generality that the sign of $a_n$ agrees with the parity of $n$. If $sum a_n$ is not absolutely convergent then $left|sum a_2nright|$ or $left|sum a_2n+1right|$ is unbounded (or both are). By 2. and 3. both $sum' a_2n$ and $sum' a_2n+1$ have to be finite. Assuming that $|sum a_2n|$ is unbounded, by 2. and 3. again
$$ sum_ngeq 1'|a_2n|geq sum_n=1^N|a_2n| $$
has to hold for any $NinmathbbN^+$, but that implies $left|sum' a_2nright|=+infty.$
answered 3 hours ago
Jack D'Aurizio
279k33272648
edited 39 secs ago
answered 3 hours ago
Jack D'Aurizio
279k33272648
answered 3 hours ago
Jack D'Aurizio
279k33272648
answered 3 hours ago
Jack D'Aurizio
279k33272648
279k33272648
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Since $mathcalP(mathbbN)$ and $mathcalP_fin(mathbbN)$ have different cardinalities ($2^aleph_0$ and $aleph_0$), I believe there is no way to avoid the RRT-lack-of-commutativity phenomenon when defining series (i.e. infinite sums) through finite sums, no matter if in the standard way, through smoothed/Cesà ro sums or whatever.
– Jack D'Aurizio
3 hours ago
I think you're pretty much stuck. There are a huge number of really general notions of summability (e.g. matrix summability methods), and for quite a few of these not only do you still run into these problems, but in fact for pretty much any of these summability methods there are many series for which most rearrangements (in the Baire category sense for certain metrics) of that series have very bad convergence properties (such as every extended real number being a subsequence limit). Or something to this effect . . .
– Dave L. Renfro
3 hours ago
You need to explain to your friend sometimes things make us unhappy and there is not much to do about them.
– Pedro Tamaroff♦
2 hours ago