Proving identity using Dirichlet L functions

Question

I'm trying to prove the following identity using Dirichlet L functions :

${\displaystyle \sum _{d\mid n}\varphi (d)=n}$

I have shown proved that the Dirichlet Series of $\varphi (n)$ equals to

${\frac {\zeta (s-1)}{\zeta (s)}}$ which might help, but couldn't figure out yet how.

Answer 1

Hint: The Dirichlet generating function of $\varphi\ast \mathbf 1$ is $\frac{\zeta(s-1)}{\zeta(s)}\zeta(s)=\zeta(s-1)$, hence we can find the answer by finding the coefficient of the d.g.f. of $\zeta(s-1)$.

Answer 2

@KemonoChen already sketched a solution with Dirichlet $L$-functions. In case you're interested in an alternative, we note the desired sum is multiplicative since $\varphi$ is, and$$p\in\Bbb P,\,k\ge0\implies\sum_{d|p^k}\varphi(d)=\sum_{j=0}^k\varphi(p^j)=1+\sum_{j=1}^k(p^j-p^{j-1})=p^k.$$

Answer 3

multiplication between Dirichlet series is equal to the Dirichlet series of Dirichlet convolution on the associated arithmetic functions, i.e. $$ \sum_{m=1}^\infty{a(m)\over m^s}\sum_{k=1}^\infty{b(k)\over k^s}=\sum_{n=1}^\infty{1\over n^s}\sum_{d|n}a(d)b\left(\frac nd\right) $$

As a result, taking the reciprocal fo a Dirichlet series is equivalent to find the Dirichlet inverse:

$$ \zeta(s)=\sum_{n=1}^\infty{1\over n^s}\iff{1\over\zeta(s)}=\sum_{n=1}^\infty{\mu(n)\over n^s} $$

Since you have proved that

$$ \sum_{n=1}^\infty{\varphi(n)\over n^s}={\zeta(s-1)\over\zeta(s)} $$

we can obtain

$$ \varphi(n)=n\sum_{d|n}{\mu(d)\over d} $$

which, by Mobius inversion, implies

$$ n=\sum_{d|n}\varphi(n) $$