I want to prove that $$ \frac{1}{\zeta(s)}=\sum_{n=1}^\infty \frac{\mu(n)}{n^s}.$$
I know that the standard proof works with the Euler product formula $$\zeta(s)=\prod_{p \ \text{prime}} \frac{1}{1-p^{-s}}$$ but I am not really used to it (because I don't understand the proof for the Euler product formula). Now I want to know if there is another proof for the identity $\frac{1}{\zeta(s)}=\sum_{n=1}^\infty \frac{\mu(n)}{n^s}$. I thought there might be a method using Möbius inversion.
On
The following proof of the reciprocal formula for the Riemann Zeta function explains more of the detail and may be helpful. This answer to Euler Product formula for Riemann zeta function proof may help explain the origin of the Euler product formula.
Assume $\mathrm{Re}(s) > 1$, which ensures the $\zeta$ series is absolutely convergent and the infinite product in Euler's product formula converges, and that $\left| 1 / n^s \right| < 1 / n < 1$ for all $n \geq 2$. Denote by $p_1 < p_2 < \cdots$ the set of all primes. For comparison purposes a heuristic proof is given below also.
From the Euler product formula we have : \begin{eqnarray} \frac{1}{\zeta(s)} & = & \prod_{p \text{ prime}} \left( 1 - \frac{1}{p^s} \right) \nonumber \\ & = & \left( 1 - \frac{1}{p_1^s} \right) \left( 1 - \frac{1}{p_2^s} \right) \cdots \label{eq:inf-prod} \tag{1} \end{eqnarray}
Imagine we could treat this infinite product in the same way as a finite product and multiply out the brackets by summing all products $q$ of terms taken one from each bracket in every possible combination (as described at beginning of this answer to Alternative proofs of Euclid-Euler theorem). Then every such product $q$ comprises either finitely or infinitely many $p_r$ terms. In the latter case each term in the product $q$ has modulus of form $1 / p_r^s$. But $\left| 1 / p_r^s \right| < 1 / p_r \leq \frac{1}{2}$, therefore intuitively, as infinitely many such terms are being multiplied, $|q| \leq \left( \frac{1}{2} \right)^{\infty} = 0$. Thus the overall contribution to the expansion's summation from such products is zero.
Consider now a product $q$ where only finitely many $p_r$'s are involved. Such $q$ has form : \begin{eqnarray*} q & = & \frac{{(-1)}^k}{{(q_1 \cdots q_k)}^s} \hspace{2em} \mbox{ where $k \geq 0$ and $q_1, \ldots, q_k$ are distinct primes} \\ & = & \frac{\mu(q_1 \cdots q_k)}{{(q_1 \cdots q_k)}^s} \end{eqnarray*}
This ranges over all possible selections of distinct primes $q_1, \ldots, q_k$ ($k \geq 0$) due to the infinite product in (\ref{eq:inf-prod}), ie. $q = \mu(n) / n^s$ ranging over all possible square-free $n \in \mathbb{N}$ (which includes $1$ when $k = 0$). Then the sum of all these $q$ is :
$$\sum_{\substack{\text{finitely many} \\ p_r \text{'s involved} }} q = \sum_{ \substack{n = 1, \\ n \text{ square-free}} }^{\infty} \frac{\mu(n)}{n^s} = \sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} $$
since $\mu(n) = 0$ for any non-square-free $n \in \mathbb{N}$, and this is the required formula for $1 / \zeta(s)$.
The proposed series is :
\begin{eqnarray} S & = & \sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} \nonumber \\ & = & \sum_{\substack{n = 1, \\ n \text{ square-free}}}^{\infty} \frac{\mu(n)}{n^s} \hspace{3em} (\mbox{since } \mu(n) = 0 \mbox{ for any non-square-free } n \in \mathbb{N}) \label{eq:dirichlet} \tag{2} \end{eqnarray}
which is the Dirichlet series over the Mobius function (absolutely convergent by comparison with the $\zeta$ series).
From the Euler product formula for $\zeta$, we have : \begin{eqnarray} \frac{1}{\zeta(s)} & = & \prod_{p \text{ prime}} \left( 1 - \frac{1}{p^s} \right) \nonumber \\ & = & \lim_{m \rightarrow \infty} Z_m, \hspace{3em} \mbox{where } Z_m = \prod_{r = 1}^m \left(1 - \frac{1}{p_r^s} \right) \label{eq:limZm} \tag{3} \\ \mbox{ie. } Z_m & = & \left( 1 - \frac{1}{p_1^s} \right) \cdots \left( 1 - \frac{1}{p_m^s} \right), \hspace{3em} (2^m \mbox{ terms on multiplying out)} \nonumber \\ \therefore Z_m & = & \sum_{\substack{ k \in \mathbb{N} \text{ square-free}, \\ \text{the primes of } k \\ \subseteq \{p_1, \ldots, p_m \} }} \frac{\mu(k)}{k^s} \label{eq:Zm} \tag{4} \end{eqnarray}
since on multiplying out to the $2^m$ terms, we have a summation of all products of terms taken one from each of the $m$ brackets in every possible combination (as described at beginning of this answer to Alternative proofs of Euclid-Euler theorem). Note the summation includes $k = 1$, with the primes of $1$ being the empty set, and the expression for $Z_m$ still makes sense for $m = 0$ where we have : $$Z_0 = \displaystyle{\prod_{\varnothing}} = \sum_{k \in \{1\}} \mu(k)/k^s = 1.$$
Every $Z_m$ is thus the sum of a finite sub-series of $S$, and the indices in $Z_m$ are enclosed telescopically within each other : $Z_0 \subsetneq Z_1 \subsetneq Z_2 \subsetneq, \cdots$. Now define a sequence of non-overlapping finite sub-series of $S$, $\{S^{(r)} : r \geq 0 \}$ by : $S^{(0)} = Z_0, S^{(1)} = Z_1 \setminus Z_0, S^{(2)} = Z_2 \setminus Z_1, \ldots, S^{(r + 1)} = Z_{r + 1} \setminus Z_r (r \geq 0)$, so : $$ S^{(m)} = Z_m \setminus Z_{m - 1} = \sum_{\substack{ k \in \mathbb{N} \text{ square-free}, \\ \text{the primes of } k \\ \subseteq \{p_1, \ldots, p_m \}, \\ \text{with $p_m$ included} }} \frac{\mu(k)}{k^s}, \hspace{2em} \forall m \geq 1 $$
ie. in the Venn diagram the sets of indices of the $S^{(m)}$ are non-overlapping concentric rings :
Or formally if $t > r$, $S^{(t)} = Z_t \setminus Z_{t - 1}$, and $S^{(r)} \subseteq Z_r \subseteq Z_{t - 1} \therefore S^{(r)} \cap S^{(t)} = \varnothing$, ie. the $S^{(r)}$ are mutually disjoint.
Now taking a square-free index $n \in \mathbb{N}$ in series $S$ of equation (\ref{eq:dirichlet}), $n$ is a product of primes (not necessarily consecutive primes) and taking $m$ sufficiently large $n$ must appear as an index in the finite summation $Z_m$ of equation (\ref{eq:Zm}). Then there is a least such $m \geq 0$. If $m = 0$, then $n$ (as an index) $\in S^{(0)}$, and $n = 1$ in this case, otherwise $n$ (as an index) $\in Z_m \setminus Z_{m - 1} = S^{(m)}$.
Thus the mutually disjoint sets of indices of the finite sub-series $S^{(r)} (r \geq 0)$ of $S$ form a partition of the set of all indices of series $S$ of equation (\ref{eq:dirichlet}). Then in the sense of [1, p142, Theorem 4] or Theorem 1 in this answer to Rearrangements of absolutely convergent series, $\{S^{(r)} : r \geq 0 \}$ is a rearrangement of series $S$ in the extended sense, so that since $S$ is absolutely convergent, we have :
\begin{equation} S = \sum_{r = 0}^{\infty} S^{(r)} \label{eq:partition} \tag{5} \end{equation}
with the convergence absolute. But the sums of the sub-series $S^{(r)}$ are : $$ S^{(0)} = Z_0, S^{(1)} = Z_1 - Z_0, \; \ldots, \; S^{(r + 1)} = Z_{r + 1} - Z_r, $$ therefore the $m$th partial sum of $S$ in equation (\ref{eq:partition}) is :
$$ S^{(0)} + S^{(1)} + \cdots + S^{(m)} = Z_0 + (Z_1 - Z_0) + \cdots + (Z_m - Z_{m - 1}) = Z_m $$
Thus : \begin{eqnarray*} S & = & \lim_{m \rightarrow \infty} Z_m \\ & = & \frac{1}{\zeta(s)} \hspace{3em} \mbox{from equation (\ref{eq:limZm})} . \end{eqnarray*}
References
[1] Konrad Knopp (1954), Theory and Application of Infinite Series, 2nd English Edition translated from 4th German Edition, Blackie & Sons.
Let $\phi$ a multiplicative function and $\phi^*$ its convolutative inverse, that is $$ \sum\limits_{n|k} \phi(n) \phi^*(n/k) = \begin{cases} 1, & n=1, \\ 0, & else.\end{cases}$$
Assuming some mild growth condition on $\phi$ and $\phi'$, then we can define $$Z(\phi,s) = \sum\limits_{n=1}^\infty \phi(n) n^{-s}$$ and $ Z(\phi^*,s)$ analogous for suffienctly large $s \gg 0$ with absolute convergence guaranteed, then for all $s$ $$Z(\phi,s) Z(\phi^*,s) =1.$$
The idea of proof is $$ Z(\phi,s) Z(\phi^*,s) = \sum_{n,m} \frac{\phi(n) \phi^*(m)}{n^s m^s}$$ and then sum over $k=nm$ first (reorder the sum by absolute convergence). $$ \dots = \sum_{k=1}^\infty k^{-s} \sum\limits_{n|k} \phi(n) \phi^*(n/k) = 1.$$ You get the identity for $s \gg 0$ Then use uniqueness of analytic continuation to make sense for the remaining $s$.