# The Euler-$\phi$ function and the Riemann Zeta function

I was reviewing a proof I came across in which it was shown that $$\sum_{n=1}^\infty \frac{\phi(n)}{n^s} = \frac{\zeta(s-1)}{\zeta(s)}.$$ The proof provided was straightforward, and reads $$\sum_{n=1}^\infty \frac{\phi(n)}{n^s} = \sum_{n=1}^\infty \frac{1}{n^s} \sum_{d|n} d \mu\left(\frac{n}{d}\right) = \sum_{n=1}^\infty \frac{n}{n^s} \sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \frac{\zeta(s-1)}{\zeta(s)}.$$ My confusion is with the statement that $$\sum_{n=1}^\infty \frac{1}{n^s} \sum_{d|n} d \mu\left(\frac{n}{d}\right) = \sum_{n=1}^\infty \frac{n}{n^s} \sum_{n=1}^\infty \frac{\mu(n)}{n^s}.$$ Could someone help shed some light on this?

This is due to Dirichlet convolution: let $$F(s) = \sum \frac{f(n)}{n^s}$$ and $$G(s) = \sum \frac{g(n)}{n^s}$$ be Dirichlet series. Then their product $$F(s) \cdot G(s)$$ equals $$\sum \frac{(f*g)(n)}{n^s}$$, where $$(f*g)(n)=\sum_{d|n} f(d)g(n/d)$$.
In your case, $$f(n)=n$$ and $$g(n)=\mu(n)$$. $$(f*g)(n)=\sum_{d|n} d\mu(n/d)$$. Writing the series on the LHS of the equation in question as $$\sum_{n=1}^\infty \dfrac{\sum_{d|n} d\mu(n/d)}{n^s},$$ we see that we are done.