Reimann zeta function












2












$begingroup$


I started solving a problem using the transformation of Reimann zeta function into this form:



enter image description here



And I searched for the methods of doing this transformation and ended up with this one :



enter image description here



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 $k geq 1$ :



enter image description here



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







share|cite|improve this question









$endgroup$








  • 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
















2












$begingroup$


I started solving a problem using the transformation of Reimann zeta function into this form:



enter image description here



And I searched for the methods of doing this transformation and ended up with this one :



enter image description here



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 $k geq 1$ :



enter image description here



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







share|cite|improve this question









$endgroup$








  • 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














2












2








2





$begingroup$


I started solving a problem using the transformation of Reimann zeta function into this form:



enter image description here



And I searched for the methods of doing this transformation and ended up with this one :



enter image description here



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 $k geq 1$ :



enter image description here



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







share|cite|improve this question









$endgroup$




I started solving a problem using the transformation of Reimann zeta function into this form:



enter image description here



And I searched for the methods of doing this transformation and ended up with this one :



enter image description here



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 $k geq 1$ :



enter image description here



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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 11 '18 at 9:52









Maths SurvivorMaths Survivor

502219




502219








  • 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




    $begingroup$
    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










1 Answer
1






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4












$begingroup$

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.






share|cite|improve this answer











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    1 Answer
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    4












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 11 '18 at 11:03

























        answered Dec 11 '18 at 10:21









        MasacrosoMasacroso

        13.1k41747




        13.1k41747






























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