How do we prove this Moebius inversion formula for infinite series From Wikipedia I read one (generalized) version of the Möbius inversion formula, which is $$g(x)=\sum_{m=1}^{\infty}\frac{f(mx)}{m^s}\quad\text{for all }x\geq1\quad\iff f(x)=\sum_{m=1}^{\infty}\frac{\mu(m)g(mx)}{m^s}\quad\text{for all }x\geq1,$$ given both sums are absolute convergent and $\mu$ is the Möbius function.  But I really do not know how to prove this using the usual Möbius inversion formula involving summing over divisors. I don't have a clue on how to deal with the infinite sum in particular.
 A: We can begin by manipulating the left hand side expression:
\begin{align*}
\sum_{m=1}^{\infty}\frac{\mu(m)g(mx)}{m^s}&=\sum_{m=1}^{\infty}\frac{\mu(m)}{m^s}\sum_{k=1}^{\infty}\frac{f(kmx)}{k^s}\\
&=\sum_{m=1}^{\infty}\sum_{k=1}^{\infty}\frac{\mu(m)f(kmx)}{(km)^s}
\end{align*}
This last sum can be described as picking a value $m$, evaluating $\mu(m)$ and them multiplying $f(kx)/k^s$ as all values $k$ that are multuples of $m$. Instead of picking a value and looking at its multiples, we can equivilantly pick a multiple and look at its divisors. Thus, we get that
\begin{align*}
\sum_{m=1}^{\infty}\frac{\mu(m)g(mx)}{m^s}&=\sum_{m=1}^{\infty}\sum_{k=1}^{\infty}\frac{\mu(m)f(kmx)}{(km)^s}\\
&=\sum_{k=1}^{\infty}\sum_{m|k}\frac{\mu(m)f(kx)}{k^s}
\end{align*}
Now, we use the fact that $\sum_{m|k}\mu(m)=0$ for all $k\geq2$ to see that only the first term survives, so
\begin{align*}
\sum_{m=1}^{\infty}\frac{\mu(m)g(mx)}{m^s}&=\sum_{k=1}^{\infty}\sum_{m|k}\frac{\mu(m)f(kx)}{k^s}\\
&=\frac{f(1\cdot x)}{1^s}\sum_{m|1}\mu(m)=f(x)
\end{align*}
and we are done.
NOTE: This trick of changing the order of summation is EXTREMELY IMPORTANT. This sort of trick is extremely common is a large amount of number theory and should be kept in the back of ones head when one is working with any sort of sum of arithmetic functions, especially the Mobius function.
