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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?

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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.

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  • $\begingroup$ I'm aware of Dirichlet convolution; I'm just confused about the final equality in my question. $\endgroup$ – user312437 Apr 7 at 23:36
  • $\begingroup$ I've added more detail. $\endgroup$ – Lukas Kofler Apr 7 at 23:41

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