Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2007 / A / 1

The Dirichlet convolution is $(f_1\ast f_2)(n)={\sum }_{d\mid n}{f}_1(d)f_2(n/d)$. Fix $s$ with $\sigma =\mathrm{Re}s>1$. Absolute convergence of the two Dirichlet series givesThus the double series may be regrouped by $n=ab$, provingThe same estimate proves absolute convergence of the resulting Dirichlet series.

For a finite set of primes $p\le y$, multiply the absolutely convergent local series. Unique prime factorization and multiplicativity, with $g(1)=1$, giveThe omitted integers all exceed $y$, so the absolute difference from $G(s)$ is at most ${\sum }_{n>y}|g(n)|n^{-\sigma }$, which tends to zero. Hence the finite products converge to $G(s)$:This proves the Euler product directly from absolute convergence, rather than treating an infinite formal multiplication as automatically valid.

The Möbius function satisfies $\mu (p^e{)}^2=1$ for $e=0,1$ and is zero for $e\ge 2$. ThereforeThe inverse local factor is $(1+p^{-s}{)}^{-1}={\sum }_{e\ge 0}(-1{)}^e{p}^{-es}$. Consequently the inverse under Dirichlet convolution is the Liouville functionwhere $\Omega (n)$ counts prime factors with multiplicity. Both functions are multiplicative functions. At $p^e$ with $e\ge 1$, their convolution is $(-1{)}^e+(-1{)}^{e-1}=0$, while its value at one is one. Multiplicativity then proves $h_1\ast h_2=I$ on every positive integer. Its Dirichlet series is $H_2(s)=\zeta (2s)/\zeta (s)$ for $\mathrm{Re}s>1$.