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







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














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

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



















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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015855%2fa-proof-of-sum-limits-dn-sigmad-n-sum-limits-dn-taud-over-d%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015855%2fa-proof-of-sum-limits-dn-sigmad-n-sum-limits-dn-taud-over-d%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Le Mesnil-Réaume

Image
Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more

web3.py web3.isConnected() returns false always

Image
1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
Read more