Mixing
Is there a comprehensive list of convergence tests for double series?
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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
MathPondererMathPonderer
243
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add a comment |
$begingroup$
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
MathPondererMathPonderer
243
$endgroup$
$begingroup$
See here math.stackexchange.com/questions/2957537/…
$endgroup$
– Nikos Bagis
Jan 10 at 6:54
add a comment |
$begingroup$
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
MathPondererMathPonderer
243
$endgroup$
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
MathPondererMathPonderer
243
asked Sep 30 '18 at 19:26
MathPondererMathPonderer
243
asked Sep 30 '18 at 19:26
MathPondererMathPonderer
243
asked Sep 30 '18 at 19:26
MathPondererMathPonderer
243
asked Sep 30 '18 at 19:26
MathPondererMathPonderer
243
243
$begingroup$
See here math.stackexchange.com/questions/2957537/…
$endgroup$
– Nikos Bagis
Jan 10 at 6:54
add a comment |
$begingroup$
See here math.stackexchange.com/questions/2957537/…
$endgroup$
– Nikos Bagis
Jan 10 at 6:54
$begingroup$
See here math.stackexchange.com/questions/2957537/…
$endgroup$
– Nikos Bagis
Jan 10 at 6:54
$begingroup$
See here math.stackexchange.com/questions/2957537/…
$endgroup$
– Nikos Bagis
Jan 10 at 6:54
add a comment |
1 Answer
1
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oldest
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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 IsraelRobert Israel
320k23209459
$endgroup$
$begingroup$
Does it get any easier if you assume absolute convergence?
$endgroup$
– Theo Bendit
Sep 30 '18 at 20:12
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@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.
$endgroup$
– Mees de Vries
Jan 3 at 16:08
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
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 IsraelRobert Israel
320k23209459
$endgroup$
$begingroup$
Does it get any easier if you assume absolute convergence?
$endgroup$
– Theo Bendit
Sep 30 '18 at 20:12
$begingroup$
@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.
$endgroup$
– Mees de Vries
Jan 3 at 16:08
add a comment |
$begingroup$
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 IsraelRobert Israel
320k23209459
$endgroup$
$begingroup$
Does it get any easier if you assume absolute convergence?
$endgroup$
– Theo Bendit
Sep 30 '18 at 20:12
$begingroup$
@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.
$endgroup$
– Mees de Vries
Jan 3 at 16:08
add a comment |
$begingroup$
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 IsraelRobert Israel
320k23209459
$endgroup$
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 IsraelRobert Israel
320k23209459
answered Sep 30 '18 at 20:05
Robert IsraelRobert Israel
320k23209459
answered Sep 30 '18 at 20:05
Robert IsraelRobert Israel
320k23209459
answered Sep 30 '18 at 20:05
Robert IsraelRobert Israel
320k23209459
320k23209459
$begingroup$
Does it get any easier if you assume absolute convergence?
$endgroup$
– Theo Bendit
Sep 30 '18 at 20:12
$begingroup$
@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.
$endgroup$
– Mees de Vries
Jan 3 at 16:08
add a comment |
$begingroup$
Does it get any easier if you assume absolute convergence?
$endgroup$
– Theo Bendit
Sep 30 '18 at 20:12
$begingroup$
@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.
$endgroup$
– Mees de Vries
Jan 3 at 16:08
$begingroup$
Does it get any easier if you assume absolute convergence?
$endgroup$
– Theo Bendit
Sep 30 '18 at 20:12
$begingroup$
Does it get any easier if you assume absolute convergence?
$endgroup$
– Theo Bendit
Sep 30 '18 at 20:12
$begingroup$
@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.
$endgroup$
– Mees de Vries
Jan 3 at 16:08
$begingroup$
@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.
$endgroup$
– Mees de Vries
Jan 3 at 16:08
add a comment |
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$begingroup$
See here math.stackexchange.com/questions/2957537/…
$endgroup$
– Nikos Bagis
Jan 10 at 6:54