What is the definition of $sum_{n = - infty}^{infty} a_n$
0
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I know that in Complex Analysis we use sums of the form $displaystylesum_{n = - infty}^{infty} a_n$
What is the actual meaning of this symbol? I expect that under some nice enough conditions, it would be equal to something like $displaystyle sum_{n=1}^infty b_n$, where $b_{2k} = a_k$ and $b_{2k+1} = a_{-k}$. But what is the actual definition?
complex-analysis analysis laurent-series
asked Dec 3 '18 at 1:22
OviOvi
12.4k1038112
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add a comment |
0
$begingroup$
I know that in Complex Analysis we use sums of the form $displaystylesum_{n = - infty}^{infty} a_n$
What is the actual meaning of this symbol? I expect that under some nice enough conditions, it would be equal to something like $displaystyle sum_{n=1}^infty b_n$, where $b_{2k} = a_k$ and $b_{2k+1} = a_{-k}$. But what is the actual definition?
complex-analysis analysis laurent-series
asked Dec 3 '18 at 1:22
OviOvi
12.4k1038112
$endgroup$
add a comment |
0
0
0
$begingroup$
I know that in Complex Analysis we use sums of the form $displaystylesum_{n = - infty}^{infty} a_n$
What is the actual meaning of this symbol? I expect that under some nice enough conditions, it would be equal to something like $displaystyle sum_{n=1}^infty b_n$, where $b_{2k} = a_k$ and $b_{2k+1} = a_{-k}$. But what is the actual definition?
complex-analysis analysis laurent-series
asked Dec 3 '18 at 1:22
OviOvi
12.4k1038112
$endgroup$
I know that in Complex Analysis we use sums of the form $displaystylesum_{n = - infty}^{infty} a_n$
What is the actual meaning of this symbol? I expect that under some nice enough conditions, it would be equal to something like $displaystyle sum_{n=1}^infty b_n$, where $b_{2k} = a_k$ and $b_{2k+1} = a_{-k}$. But what is the actual definition?
complex-analysis analysis laurent-series
complex-analysis analysis laurent-series
asked Dec 3 '18 at 1:22
OviOvi
12.4k1038112
asked Dec 3 '18 at 1:22
OviOvi
12.4k1038112
asked Dec 3 '18 at 1:22
OviOvi
12.4k1038112
asked Dec 3 '18 at 1:22
OviOvi
12.4k1038112
asked Dec 3 '18 at 1:22
OviOvi
12.4k1038112
12.4k1038112
add a comment |
add a comment |
1 Answer
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The sum is defined as $$displaystylesum_{n = - infty}^{infty} a_n:= lim_{Ntoinfty}sum_{n=0}^{N} a_n+lim_{Mtoinfty}sum_{n=1}^{M} a_{-n}$$
answered Dec 3 '18 at 1:26
greeliousgreelious
19410
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$begingroup$
In particular, we need both the positive subsequence and the negative subsequence to exist, instead of picking some Cauchy principal value.
$endgroup$
– obscurans
Dec 3 '18 at 1:41
$begingroup$
Thanks for the answer. Are there any theorems which allow you to play around with the sums, such as I did in my post?
$endgroup$
– Ovi
Dec 3 '18 at 1:42
$begingroup$
The usual stuff with rearranging two infinite sums - absolute convergence allows you to reorder terms.
$endgroup$
– obscurans
Dec 3 '18 at 1:52
$begingroup$
If you know that the sums are absolutely convergent (i.e. $sum|a_n|$ converges), then you may rearrange terms and sum in whatever manner you like, including defining the sequence $b_n$ as you did. However, if the sum is conditionally convergent (i.e. convergent but not absolutely convergent), then the Riemann series theorem states that you can rearrange the terms of the series so that the sum converges to any number you like, or diverges.
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– greelious
Dec 3 '18 at 1:54
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
$begingroup$
The sum is defined as $$displaystylesum_{n = - infty}^{infty} a_n:= lim_{Ntoinfty}sum_{n=0}^{N} a_n+lim_{Mtoinfty}sum_{n=1}^{M} a_{-n}$$
answered Dec 3 '18 at 1:26
greeliousgreelious
19410
$endgroup$
$begingroup$
In particular, we need both the positive subsequence and the negative subsequence to exist, instead of picking some Cauchy principal value.
$endgroup$
– obscurans
Dec 3 '18 at 1:41
$begingroup$
Thanks for the answer. Are there any theorems which allow you to play around with the sums, such as I did in my post?
$endgroup$
– Ovi
Dec 3 '18 at 1:42
$begingroup$
The usual stuff with rearranging two infinite sums - absolute convergence allows you to reorder terms.
$endgroup$
– obscurans
Dec 3 '18 at 1:52
$begingroup$
If you know that the sums are absolutely convergent (i.e. $sum|a_n|$ converges), then you may rearrange terms and sum in whatever manner you like, including defining the sequence $b_n$ as you did. However, if the sum is conditionally convergent (i.e. convergent but not absolutely convergent), then the Riemann series theorem states that you can rearrange the terms of the series so that the sum converges to any number you like, or diverges.
$endgroup$
– greelious
Dec 3 '18 at 1:54
add a comment |
3
$begingroup$
The sum is defined as $$displaystylesum_{n = - infty}^{infty} a_n:= lim_{Ntoinfty}sum_{n=0}^{N} a_n+lim_{Mtoinfty}sum_{n=1}^{M} a_{-n}$$
answered Dec 3 '18 at 1:26
greeliousgreelious
19410
$endgroup$
$begingroup$
In particular, we need both the positive subsequence and the negative subsequence to exist, instead of picking some Cauchy principal value.
$endgroup$
– obscurans
Dec 3 '18 at 1:41
$begingroup$
Thanks for the answer. Are there any theorems which allow you to play around with the sums, such as I did in my post?
$endgroup$
– Ovi
Dec 3 '18 at 1:42
$begingroup$
The usual stuff with rearranging two infinite sums - absolute convergence allows you to reorder terms.
$endgroup$
– obscurans
Dec 3 '18 at 1:52
$begingroup$
If you know that the sums are absolutely convergent (i.e. $sum|a_n|$ converges), then you may rearrange terms and sum in whatever manner you like, including defining the sequence $b_n$ as you did. However, if the sum is conditionally convergent (i.e. convergent but not absolutely convergent), then the Riemann series theorem states that you can rearrange the terms of the series so that the sum converges to any number you like, or diverges.
$endgroup$
– greelious
Dec 3 '18 at 1:54
add a comment |
3
3
3
$begingroup$
The sum is defined as $$displaystylesum_{n = - infty}^{infty} a_n:= lim_{Ntoinfty}sum_{n=0}^{N} a_n+lim_{Mtoinfty}sum_{n=1}^{M} a_{-n}$$
answered Dec 3 '18 at 1:26
greeliousgreelious
19410
$endgroup$
The sum is defined as $$displaystylesum_{n = - infty}^{infty} a_n:= lim_{Ntoinfty}sum_{n=0}^{N} a_n+lim_{Mtoinfty}sum_{n=1}^{M} a_{-n}$$
answered Dec 3 '18 at 1:26
greeliousgreelious
19410
answered Dec 3 '18 at 1:26
greeliousgreelious
19410
answered Dec 3 '18 at 1:26
greeliousgreelious
19410
answered Dec 3 '18 at 1:26
greeliousgreelious
19410
19410
$begingroup$
In particular, we need both the positive subsequence and the negative subsequence to exist, instead of picking some Cauchy principal value.
$endgroup$
– obscurans
Dec 3 '18 at 1:41
$begingroup$
Thanks for the answer. Are there any theorems which allow you to play around with the sums, such as I did in my post?
$endgroup$
– Ovi
Dec 3 '18 at 1:42
$begingroup$
The usual stuff with rearranging two infinite sums - absolute convergence allows you to reorder terms.
$endgroup$
– obscurans
Dec 3 '18 at 1:52
$begingroup$
If you know that the sums are absolutely convergent (i.e. $sum|a_n|$ converges), then you may rearrange terms and sum in whatever manner you like, including defining the sequence $b_n$ as you did. However, if the sum is conditionally convergent (i.e. convergent but not absolutely convergent), then the Riemann series theorem states that you can rearrange the terms of the series so that the sum converges to any number you like, or diverges.
$endgroup$
– greelious
Dec 3 '18 at 1:54
add a comment |
$begingroup$
In particular, we need both the positive subsequence and the negative subsequence to exist, instead of picking some Cauchy principal value.
$endgroup$
– obscurans
Dec 3 '18 at 1:41
$begingroup$
Thanks for the answer. Are there any theorems which allow you to play around with the sums, such as I did in my post?
$endgroup$
– Ovi
Dec 3 '18 at 1:42
$begingroup$
The usual stuff with rearranging two infinite sums - absolute convergence allows you to reorder terms.
$endgroup$
– obscurans
Dec 3 '18 at 1:52
$begingroup$
If you know that the sums are absolutely convergent (i.e. $sum|a_n|$ converges), then you may rearrange terms and sum in whatever manner you like, including defining the sequence $b_n$ as you did. However, if the sum is conditionally convergent (i.e. convergent but not absolutely convergent), then the Riemann series theorem states that you can rearrange the terms of the series so that the sum converges to any number you like, or diverges.
$endgroup$
– greelious
Dec 3 '18 at 1:54
$begingroup$
In particular, we need both the positive subsequence and the negative subsequence to exist, instead of picking some Cauchy principal value.
$endgroup$
– obscurans
Dec 3 '18 at 1:41
$begingroup$
In particular, we need both the positive subsequence and the negative subsequence to exist, instead of picking some Cauchy principal value.
$endgroup$
– obscurans
Dec 3 '18 at 1:41
$begingroup$
Thanks for the answer. Are there any theorems which allow you to play around with the sums, such as I did in my post?
$endgroup$
– Ovi
Dec 3 '18 at 1:42
$begingroup$
Thanks for the answer. Are there any theorems which allow you to play around with the sums, such as I did in my post?
$endgroup$
– Ovi
Dec 3 '18 at 1:42
$begingroup$
The usual stuff with rearranging two infinite sums - absolute convergence allows you to reorder terms.
$endgroup$
– obscurans
Dec 3 '18 at 1:52
$begingroup$
The usual stuff with rearranging two infinite sums - absolute convergence allows you to reorder terms.
$endgroup$
– obscurans
Dec 3 '18 at 1:52
$begingroup$
If you know that the sums are absolutely convergent (i.e. $sum|a_n|$ converges), then you may rearrange terms and sum in whatever manner you like, including defining the sequence $b_n$ as you did. However, if the sum is conditionally convergent (i.e. convergent but not absolutely convergent), then the Riemann series theorem states that you can rearrange the terms of the series so that the sum converges to any number you like, or diverges.
$endgroup$
– greelious
Dec 3 '18 at 1:54
$begingroup$
If you know that the sums are absolutely convergent (i.e. $sum|a_n|$ converges), then you may rearrange terms and sum in whatever manner you like, including defining the sequence $b_n$ as you did. However, if the sum is conditionally convergent (i.e. convergent but not absolutely convergent), then the Riemann series theorem states that you can rearrange the terms of the series so that the sum converges to any number you like, or diverges.
$endgroup$
– greelious
Dec 3 '18 at 1:54
add a comment |
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