Euler product

An Euler product factors a Dirichlet series into local factors indexed by prime numbers. For a multiplicative arithmetic function $f$ and in a half-plane of absolute convergence,

For $\sigma >1$, logarithmic expansion of the Euler product reduces positivity to $3+4\cos \theta +\cos 2\theta =2(1+\cos \theta {)}^2$. At primes dividing the modulus the zeta contribution is positive by itself. If the last factor is bounded at the line-one point, a zero of the middle factor would outweigh the zeta pole and force this product to zero. This proves the relevant nonvanishing of Dirichlet L-functions on the line one.

For a finite set of primes choose $0<g(p)<1$, put $k(p)=g(p)/(1-g(p))$, and let $W={\prod }_p(1+k(p))$. Give each squarefree integer $d$ on these primes probability $k(d)/W$. Inclusion of each prime is represented by a Bernoulli random variable with parameter $g(p)$; these are independent random variables, so $\mathbb{E}[\log d]={\sum }_{p}g(p)\log p$. If this is at most $\theta \log L$ with $0<\theta <1$, the Markov inequality givesThis turns a full Euler product into a lower bound for the truncated normalizing sum of a Selberg sieve.

For a finite set of primes,expands as the sum of $1/n$ over positive integers whose prime factors are at most $x$. It contains every term with $n\le x$, so these partial products dominate the harmonic series and diverge. Sinceuniformly for $0\le t\le 1/2$, convergence of ${\sum }_{p}1/p$ would force convergence of the logarithms of these products, a contradiction.

For $x\ge 2$, the harmonic number $H_{\lfloor x\rfloor }$ exceeds $\log x$. The finite Euler product over the primes at most $x$ contains every summand of this harmonic number. Taking logarithms and applying the geometric-series bound for the logarithmic remainder bounds the total remainder by $\frac{1}{2}{\sum }_{n=2}^{\infty }[n(n-1){]}^{-1}=1/2$, proving the displayed bound.

For every complex number $z$ with $|z|\le 1$,Applying this to each prime-power term in logarithms of Euler products proves thatfor $\sigma >1$ and every completely multiplicative $f$ bounded by one.