If $f$ is a <multiplicative arithmetic function>[multiplicative function] with $|f(n)|\leq1$, then for $\Re s>1$ its <Dirichlet series> has the <Euler product>
$$
\sum_{n=1}^{\infty}\frac{f(n)}{n^s}
=\prod_p\left(1+\frac{f(p)}{p^s}+\frac{f(p^2)}{p^{2s}}+\cdots\right).
$$
Indeed, expanding the product over a finite set of primes and using <unique prime factorization> gives the sum over integers having no other prime factors. Moreover,
$$
\sum_{n\geq1}\left|\frac{f(n)}{n^s}\right|
\leq\sum_{n\geq1}n^{-\Re s}<\infty,
$$
so <absolute convergence> permits rearrangement and passage to the limit over all primes. This proves the formula.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 150 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-150/2/a/solution