For $\sigma>1$, both relevant <Dirichlet series> converge absolutely, so their product may be rearranged:
$$
\zeta(s)\sum_{n=1}^\infty\frac{\mu(n)}{n^s}
=\sum_{n=1}^\infty\frac1{n^s}\sum_{d\mid n}\mu(d).
$$
The inner sum is one when $n=1$ and zero otherwise by <Möbius inversion>. Hence
$$
\boxed{\frac1{\zeta(s)}
=\sum_{n=1}^\infty\frac{\mu(n)}{n^s}\qquad(\sigma>1).}
$$
This is also the reciprocal of the absolutely convergent <Euler product>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-150/3/b/solution