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