How to prove that $\sigma_{k+1}(n) = \sum\limits_{d|n}d^k \cdot\phi(d)\cdot\sigma_{k}(\frac{n}{d})$

Question

How to prove that $\sigma_{k+1}(n) = \sum\limits_{d|n}d^k \cdot\phi(d)\cdot\sigma_{k}(\frac{n}{d})$

I've tried using Dirichlet's convolution but the $d^k$ term seems to be something I can't resolve.

Answer 1

As @reuns mentioned, it is possible to prove it using Dirichlet series. First, we have $$ \zeta(s)\zeta(s-k) = \sum_{n\geq 1} \frac{\sigma_{k}(n)}{n^{s}} $$ This follows from direct computation:

$$ \zeta(s)\zeta(s-k) = \sum_{n, m\geq 1} \frac{1}{n^{s}m^{s-k}} = \sum_{n, m\geq 1} \frac{m^{k}}{(nm)^{s}}=\sum_{n\geq 1}\sum_{d|n} \frac{d^{k}}{n^{s}} = \sum_{n\geq 1} \frac{\sigma_{k}(n)}{n^{s}} $$ (of course, one need to consider convergence issues, but this can be resolved once we only consider $s$ with sufficiently large $\Re s$.) Also, we have $$ \sum_{n\geq 1} \frac{\phi(n)}{n^{s}} = \frac{\zeta(s-1)}{\zeta(s)} $$ which follows from $\sum_{d|n} \phi(d) =n$.

Then, using Dirichlet convolution, the equation just follows from $$ \zeta(s)\zeta(s-k-1) = \frac{\zeta(s-k-1)}{\zeta(s-k)}\cdot \zeta(s)\zeta(s-k) $$