Möbius inversion of $1/n$ What is $$\mu * f?$$
How can it be found?


*

*$f(n) = 1/n$.

 A: Hint: Consider $\mu*f$ at the prime powers.  $$\mu*f(p)=\frac{1}{p}-1=\frac{1-p}{p}$$
$$\mu*f(p^\alpha)=\frac{1}{p^\alpha}-\frac{1}{p^{\alpha-1}}=\frac{1-p}{p^\alpha}.$$Then if $n=p_1^{r_1}\cdots p_m^{r_m}$ we have that $$\mu*f(n)=\frac{1}{n}\prod_{p|n} \left(1-p\right).$$ Now, be expanding the product, you can write write $$\frac{1}{n}\prod_{p|n} \left(1-p\right)=\frac{1}{n}\sum_{d|n} \mu(d) d.$$  I don't think it can be simplified further.
Hope that helps,
A: $$(\mu * f)(n)=\sum_{d|n} \mu(\frac{n}{d})f(d)=$$
$$\frac{1}{n} \sum_{d|n}\mu(\frac{n}{d})\frac{n}{d}=$$
$$\frac{1}{n} \sum_{d|n} \mu(d) d$$
When $n=p^k$, then $$(\mu*f)(n) = \frac{1-p}{p^k}$$
So if $n=p_1^{k_1}p_2^{k_2}...p_t^{k_t}$, then:
$$(\mu*f)(n) = \frac{(1-p_1)(1-p_2)...(1-p_t)}{n}$$
A: Typically, when $f \colon \mathbb{N} \to \mathbb{C}$ is an arithmetic function, Möbius inversion is defined as the Dirichlet convolution
\begin{align}
(\mu * f)(n) = \sum_{d \mid n} \mu(d) f(\tfrac{n}{d}).
\end{align}
The arithmetic function $g = \mu * f$ is said to be the Möbius inverse of $f$. For your example,
\begin{align}
(\mu * f)(n) = \frac{1}{n} \sum_{d \mid n} \mu(d) d.
\end{align}
