Dtyjlui

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?

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 $$sum_{dmid n}sigma(d)=sum_{dmid n}sum_{tmid d}t=sum_{tmid n}sum_{kmid frac{n}{t}}t=sum_{tmid n}tsum_{kmid frac{n}t}1,$$
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 $delta =frac{n}{t}$.

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

We obtain
begin{align*}
color{blue}{sum_{dmid n}sigma(d)}&=sum_{dmid n}sum_{tmid d}ttag{1}\
&=sum_{{tmid dmid n}atop{t,dgeq 1}}ttag{2}\
&=sum_{{t mid tk mid n}atop{t,kgeq 1}}ttag{3}\
&=sum_{{t mid n,k mid frac{n}{t}}atop{t,kgeq 1}}ttag{4}\
&,,color{blue}{=sum_{t mid n}tsum_{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 $t|d Longleftrightarrow exists kgeq 1 : tk=d$, assuming $t,d,k$ are positive integers.




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




• 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=(ustar u)(n)\
sigma(n)&=sum_{d|n}d=(ustar N)(n)
end{align*}

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

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 $delta =frac{n}{t}$.