For any positive integer $n$, show that $\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d)$ For any positive integer $n$, show that $\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d)$

My try :
Left hand side :
$\begin{align} \sum_{d|p^k}\sigma (d) &= \sigma(p^0) + \sigma(p^1) + \sigma(p^2) + \cdots + \sigma(p^k) \\ &= \dfrac{p^{0+1}-1}{p-1} +  \dfrac{p^{1+1}-1}{p-1}  + \cdots +  \dfrac{p^{k+1}-1}{p-1}  \\&= \dfrac{1}{p-1}\left( (p + p^2 + \cdots + p^{k+1}) - (k+1)\right) \\&= \dfrac{1}{p-1}\left(\dfrac{p(p^{k+1}-1)}{p-1}- (k+1)\right) \\
\end{align}$
I'm not sure if I am in right path; this doesn't look simple. Any help ?
 A: Define
the functions $N$ and $1$
by
$N(n)=n$ and $1(n)=1$
for all $n \in \mathbb{N}.$
An equivalent, but easier to write, solution is to note that the
function 
$\sum_{d|n}\sigma(d)$
is the Dirichlet convolution 
$\sigma\ast 1$
and thus the
definition $\sigma=N*1$
implies that 
$$\sigma*1=(N*1)*1=N*(1*1).$$
Now notice that $1*1$ is the classical divisor function $\tau$
and we therefore get
$$
\sigma*1=N*\tau,$$
which provides the required answer.
A: You can reduce to prime powers to get the job done, but that doesn't get at the heart of the equality.
$$\sum_{d\mid n}\sigma(d)=\sum_{d\mid n}\sum_{r\mid d}r=\sum_{r\mid d\mid n}r=\cdots$$
Can you continue? For each $r\mid n$, how many times is $r$ a summand of the above sum?
A: looking at this intuitively, firstly we note that:
$$
\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d) \\
=  \sum_{d|n} d\tau(n/d)
$$
so now we are summing the divisors $d$ of $n$, each divisor being counted with multiplicity $\tau(n/d)$. so you just have to persuade yourself that this multiplicity is appropriate.
but this is evident if we look at a particular $d$, since for  any integer  $m$ we have 
$$
dm | n  \Leftarrow \Rightarrow m | n/d
$$
in words there is a 1-1 correspondence between (a)  multiples of $d$ which divide $n$, and (b)  divisors of $n/d$ 
