Alternate method to derive the following equation

Question

$$\sum\limits_{d\mid n} {\tau (d)\varphi (n/d) = \sigma (n)}$$

I have seen a derivation based on Dirichlet convolution: Relation between $\sigma (N)$, $\tau (N)$, and $\varphi (N)$

Is there another way to derive this without using convolutions?

Answer 1

Not really different, but you may directly use Dirichlet series. Quite obviously we have $$ \sum_{n\geq 1}\frac{\tau(n)}{n^s} = \left(\sum_{n\geq 1}\frac{1}{n^s}\right)^2 = \zeta(s)^2 $$ for any $s$ with a sufficiently large real part. Similarly $$ \sum_{n\geq 1}\frac{\sigma(n)}{n^s}=\sum_{n\geq 1}\frac{n}{n^s}\sum_{n\geq 1}\frac{1}{n^s} = \zeta(s-1)\zeta(s), $$ so in order to get that $\tau * \varphi = \sigma $ it is enough to check that the Dirichlet series associated to $\varphi$ is $\frac{\zeta(s-1)}{\zeta(s)}$.
On the other hand, by Euler's product, $$ \sum_{n\geq 1}\frac{\varphi(n)}{n^s}=\prod_{p}\left(1+\frac{\varphi(p)}{p^s}+\frac{\varphi(p^2)}{p^{2s}}+\ldots\right)=\prod_{p}\left(1+\frac{p-1}{p^s}+\frac{p^2-p}{p^{2s}}+\ldots\right) $$ or $$ \sum_{n\geq 1}\frac{\varphi(n)}{n^s}=\prod_{p}\frac{p^s-1}{p^s-p}=\prod_{p}\frac{1-\frac{1}{p^s}}{1-\frac{1}{p^{s-1}}}=\frac{\zeta(s-1)}{\zeta(s)}, $$ done.