A proof of $sumlimits_{d|n} sigma(d) = n sumlimits_{d|n} {tau(d) over d}$












4














I'm trying to proof the following statement:



Let $n in mathbb{Z}$ and the $sum$ are on the divisors $d$ of $n$. Show that
$$sumlimits_{d|n} sigma(d) = n sumlimits_{d|n} {tau(d) over d}.$$



I arrive to a point where I have a product of polynomials and I can't simply find a way to reorder the factors in order to resemble some kind of useful structure for this problem. Any suggestions?






elementary-number-theory summation divisor-sum arithmetic-functions divisor-counting-function







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  • Use math.stackexchange.com/questions/1948422/…
    – lab bhattacharjee
    Nov 27 at 14:51






  • 1




    The sums express two different ways to count the sum of how many divisors $cmid n$ each divisor $dmid n$ divides.
    – Servaes
    Nov 27 at 14:55
















4














I'm trying to proof the following statement:



Let $n in mathbb{Z}$ and the $sum$ are on the divisors $d$ of $n$. Show that
$$sumlimits_{d|n} sigma(d) = n sumlimits_{d|n} {tau(d) over d}.$$



I arrive to a point where I have a product of polynomials and I can't simply find a way to reorder the factors in order to resemble some kind of useful structure for this problem. Any suggestions?






elementary-number-theory summation divisor-sum arithmetic-functions divisor-counting-function







share|cite|improve this question
























  • Use math.stackexchange.com/questions/1948422/…
    – lab bhattacharjee
    Nov 27 at 14:51






  • 1




    The sums express two different ways to count the sum of how many divisors $cmid n$ each divisor $dmid n$ divides.
    – Servaes
    Nov 27 at 14:55














4












4








4


2





I'm trying to proof the following statement:



Let $n in mathbb{Z}$ and the $sum$ are on the divisors $d$ of $n$. Show that
$$sumlimits_{d|n} sigma(d) = n sumlimits_{d|n} {tau(d) over d}.$$



I arrive to a point where I have a product of polynomials and I can't simply find a way to reorder the factors in order to resemble some kind of useful structure for this problem. Any suggestions?






elementary-number-theory summation divisor-sum arithmetic-functions divisor-counting-function







share|cite|improve this question















I'm trying to proof the following statement:



Let $n in mathbb{Z}$ and the $sum$ are on the divisors $d$ of $n$. Show that
$$sumlimits_{d|n} sigma(d) = n sumlimits_{d|n} {tau(d) over d}.$$



I arrive to a point where I have a product of polynomials and I can't simply find a way to reorder the factors in order to resemble some kind of useful structure for this problem. Any suggestions?






elementary-number-theory summation divisor-sum arithmetic-functions divisor-counting-function




elementary-number-theory summation divisor-sum arithmetic-functions divisor-counting-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 at 14:53









Zvi

4,670430




4,670430










asked Nov 27 at 14:46









Alessar

20913




20913












  • Use math.stackexchange.com/questions/1948422/…
    – lab bhattacharjee
    Nov 27 at 14:51






  • 1




    The sums express two different ways to count the sum of how many divisors $cmid n$ each divisor $dmid n$ divides.
    – Servaes
    Nov 27 at 14:55


















  • Use math.stackexchange.com/questions/1948422/…
    – lab bhattacharjee
    Nov 27 at 14:51






  • 1




    The sums express two different ways to count the sum of how many divisors $cmid n$ each divisor $dmid n$ divides.
    – Servaes
    Nov 27 at 14:55
















Use math.stackexchange.com/questions/1948422/…
– lab bhattacharjee
Nov 27 at 14:51




Use math.stackexchange.com/questions/1948422/…
– lab bhattacharjee
Nov 27 at 14:51




1




1




The sums express two different ways to count the sum of how many divisors $cmid n$ each divisor $dmid n$ divides.
– Servaes
Nov 27 at 14:55




The sums express two different ways to count the sum of how many divisors $cmid n$ each divisor $dmid n$ divides.
– Servaes
Nov 27 at 14:55










2 Answers
2






active

oldest

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3














We have $$sum_{dmid n}sigma(d)=sum_{dmid n}sum_{tmid d}t=sum_{tmid n}sum_{kmid frac{n}{t}}t=sum_{tmid n}tsum_{kmid frac{n}t}1,$$
where $k=frac{d}{t}$. So
$$sum_{dmid n}sigma(d)=sum_{tmid n}ttauleft(frac{n}tright)=sum_{delta mid n}frac{n}{delta}tau(delta)=nsum_{deltamid n}frac{tau(delta)}{delta},$$
where $delta =frac{n}{t}$.






share|cite|improve this answer





















  • In the final result $delta$ has the same exact meaning of $d$, identify the same divisors, right?
    – Alessar
    Nov 27 at 15:01






  • 1




    Yep. I used a different variable because I wanted to define $delta=frac{n}{t}$, which could be confusing if I say $d=frac{n}{t}$, since $d$ is used in $k=frac{d}{t}$ already. But then, which ever variables you are using as the indices of summation don't really matter.
    – Zvi
    Nov 27 at 15:03












  • Thank you so much, I really need to handle this kind of manipulations, now this is clear
    – Alessar
    Nov 27 at 15:04



















2














By way of enrichment, treating Dirichlet series as formal power series
we have that



$$sum_{qge 0} frac{sigma(q)}{q^s}
= zeta(s) sum_{qge 0} frac{q}{q^s} = zeta(s) zeta(s-1)
quadtext{so that}quad
sum_{qge 0} frac{1}{q^s} sum_{d|q} sigma(d)
= zeta(s)^2 zeta(s-1).$$



On the other hand



$$sum_{qge 0} frac{tau(q)}{q^s}
= zeta(s)^2
quadtext{so that}quad
sum_{qge 0} frac{tau(q)/q}{q^s} = zeta(s+1)^2$$



and



$$sum_{qge 0} frac{1}{q^s} sum_{d|q} frac{tau(d)}{d}
= zeta(s+1)^2 zeta(s)
quadtext{so that}quad
sum_{qge 0} frac{1}{q^s} q sum_{d|q} frac{tau(d)}{d}
= zeta(s)^2 zeta(s-1).$$



We have equality, which is the claim. This holds formally as well as
by an inverse Mellin transform in the appropriate half-plane, which in
our case is $Re(s) gt 2.$






share|cite|improve this answer





















  • Thanks for this proof, it's very detailed and interesting!
    – Alessar
    Nov 28 at 15:43











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














We have $$sum_{dmid n}sigma(d)=sum_{dmid n}sum_{tmid d}t=sum_{tmid n}sum_{kmid frac{n}{t}}t=sum_{tmid n}tsum_{kmid frac{n}t}1,$$
where $k=frac{d}{t}$. So
$$sum_{dmid n}sigma(d)=sum_{tmid n}ttauleft(frac{n}tright)=sum_{delta mid n}frac{n}{delta}tau(delta)=nsum_{deltamid n}frac{tau(delta)}{delta},$$
where $delta =frac{n}{t}$.






share|cite|improve this answer





















  • In the final result $delta$ has the same exact meaning of $d$, identify the same divisors, right?
    – Alessar
    Nov 27 at 15:01






  • 1




    Yep. I used a different variable because I wanted to define $delta=frac{n}{t}$, which could be confusing if I say $d=frac{n}{t}$, since $d$ is used in $k=frac{d}{t}$ already. But then, which ever variables you are using as the indices of summation don't really matter.
    – Zvi
    Nov 27 at 15:03












  • Thank you so much, I really need to handle this kind of manipulations, now this is clear
    – Alessar
    Nov 27 at 15:04
















3














We have $$sum_{dmid n}sigma(d)=sum_{dmid n}sum_{tmid d}t=sum_{tmid n}sum_{kmid frac{n}{t}}t=sum_{tmid n}tsum_{kmid frac{n}t}1,$$
where $k=frac{d}{t}$. So
$$sum_{dmid n}sigma(d)=sum_{tmid n}ttauleft(frac{n}tright)=sum_{delta mid n}frac{n}{delta}tau(delta)=nsum_{deltamid n}frac{tau(delta)}{delta},$$
where $delta =frac{n}{t}$.






share|cite|improve this answer





















  • In the final result $delta$ has the same exact meaning of $d$, identify the same divisors, right?
    – Alessar
    Nov 27 at 15:01






  • 1




    Yep. I used a different variable because I wanted to define $delta=frac{n}{t}$, which could be confusing if I say $d=frac{n}{t}$, since $d$ is used in $k=frac{d}{t}$ already. But then, which ever variables you are using as the indices of summation don't really matter.
    – Zvi
    Nov 27 at 15:03












  • Thank you so much, I really need to handle this kind of manipulations, now this is clear
    – Alessar
    Nov 27 at 15:04














3












3








3






We have $$sum_{dmid n}sigma(d)=sum_{dmid n}sum_{tmid d}t=sum_{tmid n}sum_{kmid frac{n}{t}}t=sum_{tmid n}tsum_{kmid frac{n}t}1,$$
where $k=frac{d}{t}$. So
$$sum_{dmid n}sigma(d)=sum_{tmid n}ttauleft(frac{n}tright)=sum_{delta mid n}frac{n}{delta}tau(delta)=nsum_{deltamid n}frac{tau(delta)}{delta},$$
where $delta =frac{n}{t}$.






share|cite|improve this answer












We have $$sum_{dmid n}sigma(d)=sum_{dmid n}sum_{tmid d}t=sum_{tmid n}sum_{kmid frac{n}{t}}t=sum_{tmid n}tsum_{kmid frac{n}t}1,$$
where $k=frac{d}{t}$. So
$$sum_{dmid n}sigma(d)=sum_{tmid n}ttauleft(frac{n}tright)=sum_{delta mid n}frac{n}{delta}tau(delta)=nsum_{deltamid n}frac{tau(delta)}{delta},$$
where $delta =frac{n}{t}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 at 14:51









Zvi

4,670430




4,670430












  • In the final result $delta$ has the same exact meaning of $d$, identify the same divisors, right?
    – Alessar
    Nov 27 at 15:01






  • 1




    Yep. I used a different variable because I wanted to define $delta=frac{n}{t}$, which could be confusing if I say $d=frac{n}{t}$, since $d$ is used in $k=frac{d}{t}$ already. But then, which ever variables you are using as the indices of summation don't really matter.
    – Zvi
    Nov 27 at 15:03












  • Thank you so much, I really need to handle this kind of manipulations, now this is clear
    – Alessar
    Nov 27 at 15:04


















  • In the final result $delta$ has the same exact meaning of $d$, identify the same divisors, right?
    – Alessar
    Nov 27 at 15:01






  • 1




    Yep. I used a different variable because I wanted to define $delta=frac{n}{t}$, which could be confusing if I say $d=frac{n}{t}$, since $d$ is used in $k=frac{d}{t}$ already. But then, which ever variables you are using as the indices of summation don't really matter.
    – Zvi
    Nov 27 at 15:03












  • Thank you so much, I really need to handle this kind of manipulations, now this is clear
    – Alessar
    Nov 27 at 15:04
















In the final result $delta$ has the same exact meaning of $d$, identify the same divisors, right?
– Alessar
Nov 27 at 15:01




In the final result $delta$ has the same exact meaning of $d$, identify the same divisors, right?
– Alessar
Nov 27 at 15:01




1




1




Yep. I used a different variable because I wanted to define $delta=frac{n}{t}$, which could be confusing if I say $d=frac{n}{t}$, since $d$ is used in $k=frac{d}{t}$ already. But then, which ever variables you are using as the indices of summation don't really matter.
– Zvi
Nov 27 at 15:03






Yep. I used a different variable because I wanted to define $delta=frac{n}{t}$, which could be confusing if I say $d=frac{n}{t}$, since $d$ is used in $k=frac{d}{t}$ already. But then, which ever variables you are using as the indices of summation don't really matter.
– Zvi
Nov 27 at 15:03














Thank you so much, I really need to handle this kind of manipulations, now this is clear
– Alessar
Nov 27 at 15:04




Thank you so much, I really need to handle this kind of manipulations, now this is clear
– Alessar
Nov 27 at 15:04











2














By way of enrichment, treating Dirichlet series as formal power series
we have that



$$sum_{qge 0} frac{sigma(q)}{q^s}
= zeta(s) sum_{qge 0} frac{q}{q^s} = zeta(s) zeta(s-1)
quadtext{so that}quad
sum_{qge 0} frac{1}{q^s} sum_{d|q} sigma(d)
= zeta(s)^2 zeta(s-1).$$



On the other hand



$$sum_{qge 0} frac{tau(q)}{q^s}
= zeta(s)^2
quadtext{so that}quad
sum_{qge 0} frac{tau(q)/q}{q^s} = zeta(s+1)^2$$



and



$$sum_{qge 0} frac{1}{q^s} sum_{d|q} frac{tau(d)}{d}
= zeta(s+1)^2 zeta(s)
quadtext{so that}quad
sum_{qge 0} frac{1}{q^s} q sum_{d|q} frac{tau(d)}{d}
= zeta(s)^2 zeta(s-1).$$



We have equality, which is the claim. This holds formally as well as
by an inverse Mellin transform in the appropriate half-plane, which in
our case is $Re(s) gt 2.$






share|cite|improve this answer





















  • Thanks for this proof, it's very detailed and interesting!
    – Alessar
    Nov 28 at 15:43
















2














By way of enrichment, treating Dirichlet series as formal power series
we have that



$$sum_{qge 0} frac{sigma(q)}{q^s}
= zeta(s) sum_{qge 0} frac{q}{q^s} = zeta(s) zeta(s-1)
quadtext{so that}quad
sum_{qge 0} frac{1}{q^s} sum_{d|q} sigma(d)
= zeta(s)^2 zeta(s-1).$$



On the other hand



$$sum_{qge 0} frac{tau(q)}{q^s}
= zeta(s)^2
quadtext{so that}quad
sum_{qge 0} frac{tau(q)/q}{q^s} = zeta(s+1)^2$$



and



$$sum_{qge 0} frac{1}{q^s} sum_{d|q} frac{tau(d)}{d}
= zeta(s+1)^2 zeta(s)
quadtext{so that}quad
sum_{qge 0} frac{1}{q^s} q sum_{d|q} frac{tau(d)}{d}
= zeta(s)^2 zeta(s-1).$$



We have equality, which is the claim. This holds formally as well as
by an inverse Mellin transform in the appropriate half-plane, which in
our case is $Re(s) gt 2.$






share|cite|improve this answer





















  • Thanks for this proof, it's very detailed and interesting!
    – Alessar
    Nov 28 at 15:43














2












2








2






By way of enrichment, treating Dirichlet series as formal power series
we have that



$$sum_{qge 0} frac{sigma(q)}{q^s}
= zeta(s) sum_{qge 0} frac{q}{q^s} = zeta(s) zeta(s-1)
quadtext{so that}quad
sum_{qge 0} frac{1}{q^s} sum_{d|q} sigma(d)
= zeta(s)^2 zeta(s-1).$$



On the other hand



$$sum_{qge 0} frac{tau(q)}{q^s}
= zeta(s)^2
quadtext{so that}quad
sum_{qge 0} frac{tau(q)/q}{q^s} = zeta(s+1)^2$$



and



$$sum_{qge 0} frac{1}{q^s} sum_{d|q} frac{tau(d)}{d}
= zeta(s+1)^2 zeta(s)
quadtext{so that}quad
sum_{qge 0} frac{1}{q^s} q sum_{d|q} frac{tau(d)}{d}
= zeta(s)^2 zeta(s-1).$$



We have equality, which is the claim. This holds formally as well as
by an inverse Mellin transform in the appropriate half-plane, which in
our case is $Re(s) gt 2.$






share|cite|improve this answer












By way of enrichment, treating Dirichlet series as formal power series
we have that



$$sum_{qge 0} frac{sigma(q)}{q^s}
= zeta(s) sum_{qge 0} frac{q}{q^s} = zeta(s) zeta(s-1)
quadtext{so that}quad
sum_{qge 0} frac{1}{q^s} sum_{d|q} sigma(d)
= zeta(s)^2 zeta(s-1).$$



On the other hand



$$sum_{qge 0} frac{tau(q)}{q^s}
= zeta(s)^2
quadtext{so that}quad
sum_{qge 0} frac{tau(q)/q}{q^s} = zeta(s+1)^2$$



and



$$sum_{qge 0} frac{1}{q^s} sum_{d|q} frac{tau(d)}{d}
= zeta(s+1)^2 zeta(s)
quadtext{so that}quad
sum_{qge 0} frac{1}{q^s} q sum_{d|q} frac{tau(d)}{d}
= zeta(s)^2 zeta(s-1).$$



We have equality, which is the claim. This holds formally as well as
by an inverse Mellin transform in the appropriate half-plane, which in
our case is $Re(s) gt 2.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 at 15:39









Marko Riedel

39.1k339107




39.1k339107












  • Thanks for this proof, it's very detailed and interesting!
    – Alessar
    Nov 28 at 15:43


















  • Thanks for this proof, it's very detailed and interesting!
    – Alessar
    Nov 28 at 15:43
















Thanks for this proof, it's very detailed and interesting!
– Alessar
Nov 28 at 15:43




Thanks for this proof, it's very detailed and interesting!
– Alessar
Nov 28 at 15:43


















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