A proof of $\sum\limits_{d|n} \sigma(d) = n \sum\limits_{d|n} {\tau(d) \over d}$

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 $$\sum\limits_{d|n} \sigma(d) = n \sum\limits_{d|n} {\tau(d) \over d}.$$

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?

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EDIT: due to unsuscribe of user593746, I can't ask delucidation, so I reopen the question because I wish to have detailed delucidation on the passages on the first row of his good proof (summatory index manipulations and t outside the first sum):

We have $$\sum_{d\mid n}\sigma(d)=\sum_{d\mid n}\sum_{t\mid d}t=\sum_{t\mid n}\sum_{k\mid \frac{n}{t}}t=\sum_{t\mid n}t\sum_{k\mid \frac{n}t}1,$$ where $k=\frac{d}{t}$. So $$\sum_{d\mid n}\sigma(d)=\sum_{t\mid n}t\tau\left(\frac{n}t\right)=\sum_{\delta \mid n}\frac{n}{\delta}\tau(\delta)=n\sum_{\delta\mid n}\frac{\tau(\delta)}{\delta},$$ where $\delta =\frac{n}{t}$.

Answer 1

Here is a derivation with some intermediate steps which might be helpful.

We obtain \begin{align*} \color{blue}{\sum_{d\mid n}\sigma(d)}&=\sum_{d\mid n}\sum_{t\mid d}t\tag{1}\\ &=\sum_{{t\mid d\mid n}\atop{t,d\geq 1}}t\tag{2}\\ &=\sum_{{t \mid tk \mid n}\atop{t,k\geq 1}}t\tag{3}\\ &=\sum_{{t \mid n,k \mid \frac{n}{t}}\atop{t,k\geq 1}}t\tag{4}\\ &\,\,\color{blue}{=\sum_{t \mid n}t\sum_{k \mid \frac{n}{t}}1}\tag{5} \end{align*}

Comment:

• In (1) we use the definition of the divisor function $\sigma(d)=\sum_{t| d}t$.

• In (2) we write the index region somewhat more compactly. This does not change the sum, as it is only a rearrangement of the summands.

• In (3) we use $t|d \Longleftrightarrow \exists k\geq 1 : tk=d$, assuming $t,d,k$ are positive integers.

• In (4) we use the transitivity of the $|$ operator: $t| d| n\Longrightarrow t|n$ and we also use $t \cdot k| n\Longleftrightarrow t | \frac{n}{k}$.

• In (5) we rearrange the summands again by summing at first over $t| n$. We also factor out $t$ from the inner sum, since $t$ does not depend on the index variable $k$.

[Add-on 2019-01-17]: A derivation using Dirichlet convolution due to OPs comment.

Taking the unit-function $u(n)=1$ for all $n$ and the function $N(n)=n$ for all $n$ we have \begin{align*} \tau(n)&=\sum_{d|n}1=(u\star u)(n)\\ \sigma(n)&=\sum_{d|n}d=(u\star N)(n) \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{d|n}\sigma(d)}&=(\sigma \star u)(n)\\ &=((u\star N)\star u)(n)\\ &=(u\star(N \star u))(n)\\ &=(u\star (u\star N))(n)\\ &=((u\star u) \star N)(n)\\ &=(\tau \star N)(n)\\ &\,\,\color{blue}{=\sum_{d|n}\tau(d)\frac{n}{d}} \end{align*} and the claim follows.

Answer 2

We have $$\sum_{d\mid n}\sigma(d)=\sum_{d\mid n}\sum_{t\mid d}t=\sum_{t\mid n}\sum_{k\mid \frac{n}{t}}t=\sum_{t\mid n}t\sum_{k\mid \frac{n}t}1,$$ where $k=\frac{d}{t}$. So $$\sum_{d\mid n}\sigma(d)=\sum_{t\mid n}t\tau\left(\frac{n}t\right)=\sum_{\delta \mid n}\frac{n}{\delta}\tau(\delta)=n\sum_{\delta\mid n}\frac{\tau(\delta)}{\delta},$$ where $\delta =\frac{n}{t}$.