We want to show \begin{align} \sum_{d|n,\ d>0}(\sigma(d)/d)\cdot \mu(n/d) =1/n , \end{align} where $\sigma(m)$ denotes the sum of all positive divisors of $m$ and where $\mu$ is the Möbius function.
The hint is that we can use the formula $\sigma(n)/n=\sum_{d|n,\ d>0} 1/d$.
I tried to pull out (1/d) from the sum and then convoluted the function to get \begin{align} \sum_{d|n,\ d>0}(\sigma(d)/d)\cdot \mu(n/d) =\sigma(n)/n\sum_{d|n,\ d>0} (\sigma(n/d))\cdot\mu(d) \end{align} and then tried to get $$\sum_{d|n,\ d>0} (\sigma(n/d))\cdot\mu(d)=1/\sigma(n)$$
But I can't figure out how to get there. Can anybody give me a hint?
Hello and welcome to MSE.
Write $ \frac{σ(n)}{n} = (U*f)(n)$ , where $f(n)=\frac{1}{n}$, and $U(n)=1$ for all n.
You are then asked to find $((U*f) * (μ))$. Dirichlet convolution is associative and commutative, therefore, $(U*f) * (μ)= f*(U*μ)$, and $U* \mu$ is the identity of the Dirichlet convolution, therefore overall we get $f(n)$, as required.
(here * means dirichlet convolution (https://en.wikipedia.org/wiki/Dirichlet_convolution))