Reimann zeta function
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I started solving a problem using the transformation of Reimann zeta function into this form:
And I searched for the methods of doing this transformation and ended up with this one :
It is a little bit unclear to me the way it exchanges the summation signs , the first summation changes variables from $n geq 1 $ to $k geq 1$ :
Any explanation to this , or any help on how to transform Reimann Zeta function into that form(using other methods) would be appreciated, thank you.
real-analysis integration
asked Dec 11 '18 at 9:52
Maths SurvivorMaths Survivor
492219
$endgroup$
add a comment |
$begingroup$
I started solving a problem using the transformation of Reimann zeta function into this form:
And I searched for the methods of doing this transformation and ended up with this one :
It is a little bit unclear to me the way it exchanges the summation signs , the first summation changes variables from $n geq 1 $ to $k geq 1$ :
Any explanation to this , or any help on how to transform Reimann Zeta function into that form(using other methods) would be appreciated, thank you.
real-analysis integration
asked Dec 11 '18 at 9:52
Maths SurvivorMaths Survivor
492219
$endgroup$
1
$begingroup$
There is nothing strange in this, it's just an interchange of the two sums. If you notice, the first one says "$n$ is greater or equal to $1$ and $k$ must be equal or greater to $n$". Then we may change it into "$k$ is greater or equal to $1$ provided that $n$ is equal or less than $k$". Simple reasoning.
$endgroup$
– Von Neumann
Dec 11 '18 at 9:55
$begingroup$
@VonNeumann so are this to expressions $sum_{n=1}^{infty} sum_{k=n}^{infty}$ and $sum_{k=1}^{infty} sum_{n=k}^{infty}$ equal , I mean am I allowed to do these kinda exchanges ?
$endgroup$
– Maths Survivor
Dec 11 '18 at 9:59
$begingroup$
You need absolute convergence of the double-series in order to do this. (This is fulfilled here.)
$endgroup$
– p4sch
Dec 11 '18 at 10:24
add a comment |
$begingroup$
I started solving a problem using the transformation of Reimann zeta function into this form:
And I searched for the methods of doing this transformation and ended up with this one :
It is a little bit unclear to me the way it exchanges the summation signs , the first summation changes variables from $n geq 1 $ to $k geq 1$ :
Any explanation to this , or any help on how to transform Reimann Zeta function into that form(using other methods) would be appreciated, thank you.
real-analysis integration
asked Dec 11 '18 at 9:52
Maths SurvivorMaths Survivor
492219
$endgroup$
I started solving a problem using the transformation of Reimann zeta function into this form:
And I searched for the methods of doing this transformation and ended up with this one :
It is a little bit unclear to me the way it exchanges the summation signs , the first summation changes variables from $n geq 1 $ to $k geq 1$ :
Any explanation to this , or any help on how to transform Reimann Zeta function into that form(using other methods) would be appreciated, thank you.
real-analysis integration
real-analysis integration
asked Dec 11 '18 at 9:52
Maths SurvivorMaths Survivor
492219
asked Dec 11 '18 at 9:52
Maths SurvivorMaths Survivor
492219
asked Dec 11 '18 at 9:52
Maths SurvivorMaths Survivor
492219
asked Dec 11 '18 at 9:52
Maths SurvivorMaths Survivor
492219
asked Dec 11 '18 at 9:52
Maths SurvivorMaths Survivor
492219
492219
1
$begingroup$
There is nothing strange in this, it's just an interchange of the two sums. If you notice, the first one says "$n$ is greater or equal to $1$ and $k$ must be equal or greater to $n$". Then we may change it into "$k$ is greater or equal to $1$ provided that $n$ is equal or less than $k$". Simple reasoning.
$endgroup$
– Von Neumann
Dec 11 '18 at 9:55
$begingroup$
@VonNeumann so are this to expressions $sum_{n=1}^{infty} sum_{k=n}^{infty}$ and $sum_{k=1}^{infty} sum_{n=k}^{infty}$ equal , I mean am I allowed to do these kinda exchanges ?
$endgroup$
– Maths Survivor
Dec 11 '18 at 9:59
$begingroup$
You need absolute convergence of the double-series in order to do this. (This is fulfilled here.)
$endgroup$
– p4sch
Dec 11 '18 at 10:24
add a comment |
1
$begingroup$
There is nothing strange in this, it's just an interchange of the two sums. If you notice, the first one says "$n$ is greater or equal to $1$ and $k$ must be equal or greater to $n$". Then we may change it into "$k$ is greater or equal to $1$ provided that $n$ is equal or less than $k$". Simple reasoning.
$endgroup$
– Von Neumann
Dec 11 '18 at 9:55
$begingroup$
@VonNeumann so are this to expressions $sum_{n=1}^{infty} sum_{k=n}^{infty}$ and $sum_{k=1}^{infty} sum_{n=k}^{infty}$ equal , I mean am I allowed to do these kinda exchanges ?
$endgroup$
– Maths Survivor
Dec 11 '18 at 9:59
$begingroup$
You need absolute convergence of the double-series in order to do this. (This is fulfilled here.)
$endgroup$
– p4sch
Dec 11 '18 at 10:24
1
1
$begingroup$
There is nothing strange in this, it's just an interchange of the two sums. If you notice, the first one says "$n$ is greater or equal to $1$ and $k$ must be equal or greater to $n$". Then we may change it into "$k$ is greater or equal to $1$ provided that $n$ is equal or less than $k$". Simple reasoning.
$endgroup$
– Von Neumann
Dec 11 '18 at 9:55
$begingroup$
There is nothing strange in this, it's just an interchange of the two sums. If you notice, the first one says "$n$ is greater or equal to $1$ and $k$ must be equal or greater to $n$". Then we may change it into "$k$ is greater or equal to $1$ provided that $n$ is equal or less than $k$". Simple reasoning.
$endgroup$
– Von Neumann
Dec 11 '18 at 9:55
$begingroup$
@VonNeumann so are this to expressions $sum_{n=1}^{infty} sum_{k=n}^{infty}$ and $sum_{k=1}^{infty} sum_{n=k}^{infty}$ equal , I mean am I allowed to do these kinda exchanges ?
$endgroup$
– Maths Survivor
Dec 11 '18 at 9:59
$begingroup$
@VonNeumann so are this to expressions $sum_{n=1}^{infty} sum_{k=n}^{infty}$ and $sum_{k=1}^{infty} sum_{n=k}^{infty}$ equal , I mean am I allowed to do these kinda exchanges ?
$endgroup$
– Maths Survivor
Dec 11 '18 at 9:59
$begingroup$
You need absolute convergence of the double-series in order to do this. (This is fulfilled here.)
$endgroup$
– p4sch
Dec 11 '18 at 10:24
$begingroup$
You need absolute convergence of the double-series in order to do this. (This is fulfilled here.)
$endgroup$
– p4sch
Dec 11 '18 at 10:24
add a comment |
1 Answer
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Something that works very nice to see the change of indexes is writing a sum like this
$$sum_{color{red}{n}ge 1}sum_{color{green}{k}ge n}=sum_{1lecolor{red}{ n}<infty}sum_{nle color{green}{k}<infty}=sum_{1le color{red}{n}le color{green}{k}<infty}$$
where I use colors to denote the variables clearly in each sum. The notation at the LHS seems a bit sloppy, it is writing $kge n$ where $k$ is the variable, not $n$. Now choosing $k$ as a variable from $1$ to infinity we get
$$sum_{1le color{red}{n}lecolor{green}{k}<infty}=sum_{1lecolor{green}{ k}leinfty}sum_{1le color{red}{n}le k}=sum_{k=1}^inftysum_{n=1}^k$$
where the last expression is equivalent to the sloppy form $sum_{kge 1}sum_{nle k}$. This reordering is justified when the double sum is summable, this happen, by example, when it is a double sum of non-negative terms or when it is absolutely convergent.
answered Dec 11 '18 at 10:21
MasacrosoMasacroso
13.1k41746
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add a comment |
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$begingroup$
Something that works very nice to see the change of indexes is writing a sum like this
$$sum_{color{red}{n}ge 1}sum_{color{green}{k}ge n}=sum_{1lecolor{red}{ n}<infty}sum_{nle color{green}{k}<infty}=sum_{1le color{red}{n}le color{green}{k}<infty}$$
where I use colors to denote the variables clearly in each sum. The notation at the LHS seems a bit sloppy, it is writing $kge n$ where $k$ is the variable, not $n$. Now choosing $k$ as a variable from $1$ to infinity we get
$$sum_{1le color{red}{n}lecolor{green}{k}<infty}=sum_{1lecolor{green}{ k}leinfty}sum_{1le color{red}{n}le k}=sum_{k=1}^inftysum_{n=1}^k$$
where the last expression is equivalent to the sloppy form $sum_{kge 1}sum_{nle k}$. This reordering is justified when the double sum is summable, this happen, by example, when it is a double sum of non-negative terms or when it is absolutely convergent.
answered Dec 11 '18 at 10:21
MasacrosoMasacroso
13.1k41746
$endgroup$
add a comment |
$begingroup$
Something that works very nice to see the change of indexes is writing a sum like this
$$sum_{color{red}{n}ge 1}sum_{color{green}{k}ge n}=sum_{1lecolor{red}{ n}<infty}sum_{nle color{green}{k}<infty}=sum_{1le color{red}{n}le color{green}{k}<infty}$$
where I use colors to denote the variables clearly in each sum. The notation at the LHS seems a bit sloppy, it is writing $kge n$ where $k$ is the variable, not $n$. Now choosing $k$ as a variable from $1$ to infinity we get
$$sum_{1le color{red}{n}lecolor{green}{k}<infty}=sum_{1lecolor{green}{ k}leinfty}sum_{1le color{red}{n}le k}=sum_{k=1}^inftysum_{n=1}^k$$
where the last expression is equivalent to the sloppy form $sum_{kge 1}sum_{nle k}$. This reordering is justified when the double sum is summable, this happen, by example, when it is a double sum of non-negative terms or when it is absolutely convergent.
answered Dec 11 '18 at 10:21
MasacrosoMasacroso
13.1k41746
$endgroup$
add a comment |
$begingroup$
Something that works very nice to see the change of indexes is writing a sum like this
$$sum_{color{red}{n}ge 1}sum_{color{green}{k}ge n}=sum_{1lecolor{red}{ n}<infty}sum_{nle color{green}{k}<infty}=sum_{1le color{red}{n}le color{green}{k}<infty}$$
where I use colors to denote the variables clearly in each sum. The notation at the LHS seems a bit sloppy, it is writing $kge n$ where $k$ is the variable, not $n$. Now choosing $k$ as a variable from $1$ to infinity we get
$$sum_{1le color{red}{n}lecolor{green}{k}<infty}=sum_{1lecolor{green}{ k}leinfty}sum_{1le color{red}{n}le k}=sum_{k=1}^inftysum_{n=1}^k$$
where the last expression is equivalent to the sloppy form $sum_{kge 1}sum_{nle k}$. This reordering is justified when the double sum is summable, this happen, by example, when it is a double sum of non-negative terms or when it is absolutely convergent.
answered Dec 11 '18 at 10:21
MasacrosoMasacroso
13.1k41746
$endgroup$
Something that works very nice to see the change of indexes is writing a sum like this
$$sum_{color{red}{n}ge 1}sum_{color{green}{k}ge n}=sum_{1lecolor{red}{ n}<infty}sum_{nle color{green}{k}<infty}=sum_{1le color{red}{n}le color{green}{k}<infty}$$
where I use colors to denote the variables clearly in each sum. The notation at the LHS seems a bit sloppy, it is writing $kge n$ where $k$ is the variable, not $n$. Now choosing $k$ as a variable from $1$ to infinity we get
$$sum_{1le color{red}{n}lecolor{green}{k}<infty}=sum_{1lecolor{green}{ k}leinfty}sum_{1le color{red}{n}le k}=sum_{k=1}^inftysum_{n=1}^k$$
where the last expression is equivalent to the sloppy form $sum_{kge 1}sum_{nle k}$. This reordering is justified when the double sum is summable, this happen, by example, when it is a double sum of non-negative terms or when it is absolutely convergent.
answered Dec 11 '18 at 10:21
MasacrosoMasacroso
13.1k41746
edited Dec 11 '18 at 11:03
answered Dec 11 '18 at 10:21
MasacrosoMasacroso
13.1k41746
answered Dec 11 '18 at 10:21
MasacrosoMasacroso
13.1k41746
answered Dec 11 '18 at 10:21
MasacrosoMasacroso
13.1k41746
13.1k41746
add a comment |
add a comment |
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1
$begingroup$
There is nothing strange in this, it's just an interchange of the two sums. If you notice, the first one says "$n$ is greater or equal to $1$ and $k$ must be equal or greater to $n$". Then we may change it into "$k$ is greater or equal to $1$ provided that $n$ is equal or less than $k$". Simple reasoning.
$endgroup$
– Von Neumann
Dec 11 '18 at 9:55
$begingroup$
@VonNeumann so are this to expressions $sum_{n=1}^{infty} sum_{k=n}^{infty}$ and $sum_{k=1}^{infty} sum_{n=k}^{infty}$ equal , I mean am I allowed to do these kinda exchanges ?
$endgroup$
– Maths Survivor
Dec 11 '18 at 9:59
$begingroup$
You need absolute convergence of the double-series in order to do this. (This is fulfilled here.)
$endgroup$
– p4sch
Dec 11 '18 at 10:24