Is there a comprehensive list of convergence tests for double series?
I’m looking for a comprehensive list of methods to check for convergence of infinite double series. Specifically conditional convergence.
sequences-and-series convergence
asked Sep 30 '18 at 19:26
MathPonderer
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This question has not received enough attention.
The existing response to this question is far from being comprehensive. I think it would be beneficial for all, if a really sincere, detailed answer, carefully explaining the current state-of-affairs as far as this question is concerned, could be obtained from this platform.
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I’m looking for a comprehensive list of methods to check for convergence of infinite double series. Specifically conditional convergence.
sequences-and-series convergence
asked Sep 30 '18 at 19:26
MathPonderer
263
This question has an open bounty worth +100
reputation from Sudeepan Datta ending in 6 days.
This question has not received enough attention.
The existing response to this question is far from being comprehensive. I think it would be beneficial for all, if a really sincere, detailed answer, carefully explaining the current state-of-affairs as far as this question is concerned, could be obtained from this platform.
add a comment |
I’m looking for a comprehensive list of methods to check for convergence of infinite double series. Specifically conditional convergence.
sequences-and-series convergence
asked Sep 30 '18 at 19:26
MathPonderer
263
I’m looking for a comprehensive list of methods to check for convergence of infinite double series. Specifically conditional convergence.
sequences-and-series convergence
sequences-and-series convergence
asked Sep 30 '18 at 19:26
MathPonderer
263
asked Sep 30 '18 at 19:26
MathPonderer
263
asked Sep 30 '18 at 19:26
MathPonderer
263
asked Sep 30 '18 at 19:26
MathPonderer
263
asked Sep 30 '18 at 19:26
MathPonderer
263
263
This question has an open bounty worth +100
reputation from Sudeepan Datta ending in 6 days.
This question has not received enough attention.
The existing response to this question is far from being comprehensive. I think it would be beneficial for all, if a really sincere, detailed answer, carefully explaining the current state-of-affairs as far as this question is concerned, could be obtained from this platform.
This question has an open bounty worth +100
reputation from Sudeepan Datta ending in 6 days.
This question has not received enough attention.
The existing response to this question is far from being comprehensive. I think it would be beneficial for all, if a really sincere, detailed answer, carefully explaining the current state-of-affairs as far as this question is concerned, could be obtained from this platform.
add a comment |
add a comment |
1 Answer
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First of all, you have to be careful with exactly what you mean by "infinite double series", because rearrangements of conditionally convergent series can diverge. I'll assume that when you talk about
convergence of a double series of the form
$$ sum_{i=1}^infty sum_{j=1}^infty a_{ij}$$
you mean that for each $i$, the series $sum_{j=1}^infty a_{ij}$ converges to a value $b_i$, and the series $sum_{i=1}^infty b_i$ converges.
The methods for proving convergence of such double series are essentially the same as those for a single series, except that it may be more difficult because you might not know $b_i$ in closed form.
answered Sep 30 '18 at 20:05
Robert Israel
318k23208457
Does it get any easier if you assume absolute convergence?
– Theo Bendit
Sep 30 '18 at 20:12
@TheoBendit, if you go looking for absolute convergence (e.g. when the terms in the sum are all positive, so that convergence is equivalent to absolute convergence) then you can start rearranging the sum in interesting ways. For example, it might be that $sum_{k = 2}^infty sum_{i = 1}^{k-1} a_{i,k-i}$ is easier to evaluate, because the finite summation has a nice closed form.
– Mees de Vries
13 hours ago
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
First of all, you have to be careful with exactly what you mean by "infinite double series", because rearrangements of conditionally convergent series can diverge. I'll assume that when you talk about
convergence of a double series of the form
$$ sum_{i=1}^infty sum_{j=1}^infty a_{ij}$$
you mean that for each $i$, the series $sum_{j=1}^infty a_{ij}$ converges to a value $b_i$, and the series $sum_{i=1}^infty b_i$ converges.
The methods for proving convergence of such double series are essentially the same as those for a single series, except that it may be more difficult because you might not know $b_i$ in closed form.
answered Sep 30 '18 at 20:05
Robert Israel
318k23208457
Does it get any easier if you assume absolute convergence?
– Theo Bendit
Sep 30 '18 at 20:12
@TheoBendit, if you go looking for absolute convergence (e.g. when the terms in the sum are all positive, so that convergence is equivalent to absolute convergence) then you can start rearranging the sum in interesting ways. For example, it might be that $sum_{k = 2}^infty sum_{i = 1}^{k-1} a_{i,k-i}$ is easier to evaluate, because the finite summation has a nice closed form.
– Mees de Vries
13 hours ago
add a comment |
First of all, you have to be careful with exactly what you mean by "infinite double series", because rearrangements of conditionally convergent series can diverge. I'll assume that when you talk about
convergence of a double series of the form
$$ sum_{i=1}^infty sum_{j=1}^infty a_{ij}$$
you mean that for each $i$, the series $sum_{j=1}^infty a_{ij}$ converges to a value $b_i$, and the series $sum_{i=1}^infty b_i$ converges.
The methods for proving convergence of such double series are essentially the same as those for a single series, except that it may be more difficult because you might not know $b_i$ in closed form.
answered Sep 30 '18 at 20:05
Robert Israel
318k23208457
Does it get any easier if you assume absolute convergence?
– Theo Bendit
Sep 30 '18 at 20:12
@TheoBendit, if you go looking for absolute convergence (e.g. when the terms in the sum are all positive, so that convergence is equivalent to absolute convergence) then you can start rearranging the sum in interesting ways. For example, it might be that $sum_{k = 2}^infty sum_{i = 1}^{k-1} a_{i,k-i}$ is easier to evaluate, because the finite summation has a nice closed form.
– Mees de Vries
13 hours ago
add a comment |
First of all, you have to be careful with exactly what you mean by "infinite double series", because rearrangements of conditionally convergent series can diverge. I'll assume that when you talk about
convergence of a double series of the form
$$ sum_{i=1}^infty sum_{j=1}^infty a_{ij}$$
you mean that for each $i$, the series $sum_{j=1}^infty a_{ij}$ converges to a value $b_i$, and the series $sum_{i=1}^infty b_i$ converges.
The methods for proving convergence of such double series are essentially the same as those for a single series, except that it may be more difficult because you might not know $b_i$ in closed form.
answered Sep 30 '18 at 20:05
Robert Israel
318k23208457
First of all, you have to be careful with exactly what you mean by "infinite double series", because rearrangements of conditionally convergent series can diverge. I'll assume that when you talk about
convergence of a double series of the form
$$ sum_{i=1}^infty sum_{j=1}^infty a_{ij}$$
you mean that for each $i$, the series $sum_{j=1}^infty a_{ij}$ converges to a value $b_i$, and the series $sum_{i=1}^infty b_i$ converges.
The methods for proving convergence of such double series are essentially the same as those for a single series, except that it may be more difficult because you might not know $b_i$ in closed form.
answered Sep 30 '18 at 20:05
Robert Israel
318k23208457
answered Sep 30 '18 at 20:05
Robert Israel
318k23208457
answered Sep 30 '18 at 20:05
Robert Israel
318k23208457
answered Sep 30 '18 at 20:05
Robert Israel
318k23208457
318k23208457
Does it get any easier if you assume absolute convergence?
– Theo Bendit
Sep 30 '18 at 20:12
@TheoBendit, if you go looking for absolute convergence (e.g. when the terms in the sum are all positive, so that convergence is equivalent to absolute convergence) then you can start rearranging the sum in interesting ways. For example, it might be that $sum_{k = 2}^infty sum_{i = 1}^{k-1} a_{i,k-i}$ is easier to evaluate, because the finite summation has a nice closed form.
– Mees de Vries
13 hours ago
add a comment |
Does it get any easier if you assume absolute convergence?
– Theo Bendit
Sep 30 '18 at 20:12
@TheoBendit, if you go looking for absolute convergence (e.g. when the terms in the sum are all positive, so that convergence is equivalent to absolute convergence) then you can start rearranging the sum in interesting ways. For example, it might be that $sum_{k = 2}^infty sum_{i = 1}^{k-1} a_{i,k-i}$ is easier to evaluate, because the finite summation has a nice closed form.
– Mees de Vries
13 hours ago
Does it get any easier if you assume absolute convergence?
– Theo Bendit
Sep 30 '18 at 20:12
Does it get any easier if you assume absolute convergence?
– Theo Bendit
Sep 30 '18 at 20:12
@TheoBendit, if you go looking for absolute convergence (e.g. when the terms in the sum are all positive, so that convergence is equivalent to absolute convergence) then you can start rearranging the sum in interesting ways. For example, it might be that $sum_{k = 2}^infty sum_{i = 1}^{k-1} a_{i,k-i}$ is easier to evaluate, because the finite summation has a nice closed form.
– Mees de Vries
13 hours ago
@TheoBendit, if you go looking for absolute convergence (e.g. when the terms in the sum are all positive, so that convergence is equivalent to absolute convergence) then you can start rearranging the sum in interesting ways. For example, it might be that $sum_{k = 2}^infty sum_{i = 1}^{k-1} a_{i,k-i}$ is easier to evaluate, because the finite summation has a nice closed form.
– Mees de Vries
13 hours ago
add a comment |
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