Dirichlet Series and Euler Products

Write ${\sigma }_0:=\mathrm{\Re }{s}_0$ and $u_n:=a_n{n}^{-s_0}$. Since $\sum u_n$ converges, its partial sums $U_N$ are bounded: $\mathrm{\mid }{U}_N\mathrm{\mid }\le B$. Put $\epsilon :={\min }_{s\in K}(\mathrm{\Re }s-{\sigma }_0)>0.$ For $s\in K$, apply [L1] to the tail with weights $b_n=n^{-(s-s_0)}$. Because $b_n-b_{n+1}=O_K(n^{-1-\epsilon })$ and $\mathrm{\mid }{b}_N\mathrm{\mid }\le N^{-\epsilon }$, there is a constant $C_K$ such that $\mid {\sum }_{n=M}^N{a}_n{n}^{-s}\mid \le C_{K}B(M^{-\epsilon }+{\sum }_{n=M}^{N-1}{n}^{-1-\epsilon })\mathrm{.}$ The right-hand side tends to $0$ uniformly in $s\in K$, so the series converges uniformly on $K$.

Each partial sum is holomorphic, being a finite linear combination of the holomorphic functions $s\mapsto n^{-s}$. Since the convergence is uniform on every compact subset of the half-plane, [L2] makes the limit holomorphic there.

Write ${\sigma }_0:=\mathrm{\Re }{s}_0$. Since $\sum \mathrm{\mid }{a}_n\mathrm{\mid }{n}^{-{\sigma }_0}$ converges, for every $s$ with $\mathrm{\Re }s\ge {\sigma }_0+\epsilon$ one has $\mathrm{\mid }{a}_n{n}^{-s}\mathrm{\mid }\le \mathrm{\mid }{a}_n\mathrm{\mid }{n}^{-{\sigma }_0-\epsilon }\le \mathrm{\mid }{a}_n\mathrm{\mid }{n}^{-{\sigma }_0}\mathrm{.}$ Thus [L1] gives absolute and locally uniform convergence of the original series on the stated closed half-plane.

For large $n$ one has $\log n\le n^{\epsilon \mathrm{/}2}$, so on the same region $\mathrm{\mid }{a}_n(\log n)n^{-s}\mathrm{\mid }\le \mathrm{\mid }{a}_n\mathrm{\mid }{n}^{-{\sigma }_0-\epsilon \mathrm{/}2}\mathrm{.}$ The majorant series on the right converges because it is termwise bounded by $\sum \mathrm{\mid }{a}_n\mathrm{\mid }{n}^{-{\sigma }_0}$. Therefore [L1] also gives local uniform convergence of the derivative series.

The partial sums are holomorphic, the derivative series converges locally uniformly, and the original series converges at every point of the half-plane from step 1.1. Hence [L2] yields termwise differentiation there.

Absolute convergence implies ordinary convergence term by term, so every half-plane counted for ${\sigma }_a$ is also counted for ${\sigma }_c$. Therefore ${\sigma }_c\le {\sigma }_a$.

Let $s_0$ be any point of convergence and write ${\sigma }_0:=\mathrm{\Re }{s}_0$. Then the terms $a_n{n}^{-s_0}$ tend to $0$, so they are bounded: $\mathrm{\mid }{a}_n\mathrm{\mid }\le C{n}^{{\sigma }_0}$ for some $C$. Hence for every $s$ with $\mathrm{\Re }s>{\sigma }_0+1$, $\mathrm{\mid }{a}_n\mathrm{\mid }{n}^{-\mathrm{\Re }s}\le C{n}^{-1-\delta }$ for some $\delta >0$, and the right-hand side is summable. Thus absolute convergence holds throughout $\mathrm{\Re }s>{\sigma }_0+1$.

Since step 1.2 applies at every point of convergence, taking infima in [L1] gives ${\sigma }_a\le {\sigma }_c+1$. Combined with step 1.1, this is the claimed gap bound.

Subtract the two series. It is enough to prove that if ${\sum }_{n\ge 1}h(n)n^{-s}=0$ for all $\mathrm{\Re }s>\sigma$ and the series converges absolutely there, then $h=0$. Assume otherwise and let $m$ be the least index with $h(m)\mathrm{\ne }0$.

For real $t>\sigma$, multiply the zero identity by $m^t$: $0=h(m)+{\sum }_{n>m}h(n){\left(\frac{m}{n}\right)}^t\mathrm{.}$ Because the original series converges absolutely at one fixed real point $t_0>\sigma$, the tail is dominated by ${\sum }_{n>m}\mathrm{\mid }h(n)\mathrm{\mid }{\left(\frac{m}{n}\right)}^{t_0},$ and for each $n>m$ the factor $(m\mathrm{/}n{)}^t$ tends to $0$ as $t\to +\mathrm{\infty }$. Hence the tail tends to $0$, so letting $t\to +\mathrm{\infty }$ yields $0=h(m)$, contradiction.

Therefore no such least $m$ exists and all coefficients agree.

Fix an integer $N\ge 1$. Extend the coefficients by $a_0:=0$, set $b_0:=0$, and put $b_n:=n^{-s}$ for $1\le n\le N$. The partial sums $S_n:={\sum }_{k=0}^n{a}_k$ in [L1] then satisfy $S_0=0$ and $S_n=A(n)$ for $1\le n\le N$. Applying the tail identity in [L1] with $p=1$ and $q=N$ gives ${\sum }_{n=1}^N{a}_n{n}^{-s}=A(N)N^{-s}+{\sum }_{n=1}^{N-1}A(n)(n^{-s}-(n+1{)}^{-s})\mathrm{.}$ Since $n^{-s}-(n+1{)}^{-s}=s{\int }_n^{n+1}{x}^{-s-1}dx,$ the finite sum is exactly $s{\int }_1^{N}A(x)x^{-s-1}dx$, because $A(x)$ is constant on each interval $[n,n+1)$.

By [L2], there are $C>0$ and $x_0\ge 1$ such that, for $x\ge x_0$, $\mathrm{\mid }A(x)x^{-s-1}\mathrm{\mid }\le C{x}^{\theta -\mathrm{\Re }s-1}\mathrm{.}$ The exponent is strictly less than $-1$, so the integral over $[x_0,\mathrm{\infty })$ converges absolutely. On $[1,x_0]$, the function $A$ is a bounded step function and $x^{-s-1}$ is continuous, so the integral there also exists. For all sufficiently large $N$, the boundary term satisfies $\mathrm{\mid }A(N)N^{-s}\mathrm{\mid }\le C{N}^{\theta -\mathrm{\Re }s}⟶0.$ Letting $N\to \mathrm{\infty }$ in step 1.1 therefore proves that the Dirichlet-series partial sums converge to the stated improper integral.

Absolute convergence makes the double series ${\sum }_{m\ge 1}{\sum }_{n\ge 1}f(m)g(n)(mn{)}^{-s}$ absolutely convergent, so its terms may be regrouped by the product $mn$. The coefficient of $k^{-s}$ in that regrouping is ${\sum }_{mn=k}f(m)g(n)={\sum }_{d\mid k}f(d)g(k\mathrm{/}d)=(f\ast g)(k)$ by [L1].

Therefore the product of the two Dirichlet series is the Dirichlet series of the convolution.

For a finite set $P$ of primes, expand ${\prod }_{p\in P}{\sum }_{k\ge 0}f(p^k)p^{-ks}\mathrm{.}$ Using multiplicativity [L1] and unique factorization [L2], this is exactly ${\sum }_{\begin{array}{c}n\ge 1\\ \text{all prime factors of }n\text{ lie in }P\end{array}}f(n)n^{-s}\mathrm{.}$

As $P$ increases, these partial Euler products exhaust the original Dirichlet series. Because $\mathrm{\Re }s\ge \sigma$ and the series $\sum \mathrm{\mid }f(n)\mathrm{\mid }{n}^{-\sigma }$ converges, the omitted tail tends to $0$ absolutely. Hence the finite prime products converge to ${\sum }_{n\ge 1}f(n)n^{-s}$.

By [L2], ${\sum }_{n\ge 1}f(n)n^{-s}={\prod }_p{\sum }_{k\ge 0}f(p^k)p^{-ks}\mathrm{.}$

By [L1], each local factor is the geometric series ${\sum }_{k\ge 0}(f(p)p^{-s}{)}^k=\frac{1}{1-f(p)p^{-s}},$ whose denominator is nonzero because absolute convergence forces $\mathrm{\mid }f(p)p^{-s}\mathrm{\mid }<1$. Substituting into step 1.1 gives the claimed Euler product.

Suppose $D$ were holomorphic on a disc centered at ${\sigma }_c$. Choose a real point ${\sigma }_1>{\sigma }_c$ inside that disc and a radius $r>{\sigma }_1-{\sigma }_c$ still contained in the disc. By [L2], for every $m\ge 0$, $\frac{(-1{)}^m}{m!}{D}^{(m)}({\sigma }_1)={\sum }_{n\ge 1}{a}_n\frac{(\log n{)}^m}{m!}{n}^{-{\sigma }_1},$ and every coefficient on the right is nonnegative.

The Taylor series of $D$ at ${\sigma }_1$ therefore has nonnegative coefficients: $D(s)={\sum }_{m\ge 0}{c}_m({\sigma }_1-s{)}^m,c_m\ge 0.$ Because the disc radius exceeds ${\sigma }_1-{\sigma }_c$, this series converges at some real point ${\sigma }_{\ast }<{\sigma }_c$. Evaluating there and using the displayed formula for $c_m$ gives ${\sum }_{n\ge 1}{a}_n{n}^{-{\sigma }_{\ast }}<\mathrm{\infty }\mathrm{.}$ So the Dirichlet series converges at ${\sigma }_{\ast }$, contradicting the definition of ${\sigma }_c$ in [L1].

Hence ${\sigma }_c$ cannot be a regular point of the holomorphic continuation: it is a singular point.

Apply [L1] to the constant function $1$: for $\mathrm{\Re }s>1$, $\zeta (s)={\prod }_p\frac{1}{1-p^{-s}}\mathrm{.}$ Taking the logarithmic derivative termwise in the absolutely convergent Euler product gives $-\frac{{\zeta }^{\mathrm{\prime }}(s)}{\zeta (s)}={\sum }_p{\sum }_{k\ge 1}(\log p)p^{-ks}\mathrm{.}$

The coefficient of $n^{-s}$ on the right is $\log p$ when $n=p^k$ is a prime power and $0$ otherwise, which is exactly $\mathrm{\Lambda }(n)$ by [L2]. Therefore $-\frac{{\zeta }^{\mathrm{\prime }}(s)}{\zeta (s)}={\sum }_{n\ge 1}\mathrm{\Lambda }(n)n^{-s}\mathrm{.}$

Write $\sigma :=\mathrm{\Re }s>1$ and choose a rational $q$ with $1<q<\sigma$. Since $\mu (n)\in \{-1,0,1\}$ by [L1], one has $\mathrm{\mid }\mu (n)n^{-s}\mathrm{\mid }\le n^{-\sigma }\le n^{-q},\mathrm{\mid }{n}^{-s}\mathrm{\mid }=n^{-\sigma }\le n^{-q}\mathrm{.}$ By [L3], both Dirichlet series therefore converge absolutely.

By [L2], $({\sum }_{n\ge 1}\mu (n)n^{-s})\left({\sum }_{n\ge 1}{n}^{-s}\right)={\sum }_{n\ge 1}(\mu \ast \mathbf{1})(n)n^{-s}\mathrm{.}$

Step 2.1 and the identity in [L1] make the right-hand side equal to $1$, so the first factor is the reciprocal of $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$.

Applying [L2] to the constant-one function gives $\zeta (s{)}^2={\sum }_{n\ge 1}(\mathbf{1}\ast \mathbf{1})(n)n^{-s}\mathrm{.}$

By [L1], the coefficient $(\mathbf{1}\ast \mathbf{1})(n)$ is exactly $\tau (n)$, so step 1.1 is the claimed identity.

By [L1] and Möbius inversion from [L2], $\phi =\mu \ast {\mathrm{id}}_1\mathrm{.}$

Write $\sigma :=\mathrm{\Re }s>2$ and choose a rational $q$ with $1<q<\sigma -1$. Since $\mu (n)\in \{-1,0,1\}$, one has $\mathrm{\mid }\mu (n)n^{-s}\mathrm{\mid }\le n^{-\sigma }\le n^{-q},\mathrm{\mid }{n}^{1-s}\mathrm{\mid }=n^{1-\sigma }\le n^{-q}\mathrm{.}$ By [L4], both Dirichlet series in the next step converge absolutely.

Therefore Multiplying absolutely convergent Dirichlet series gives Dirichlet convolution gives ${\sum }_{n\ge 1}\phi (n)n^{-s}=({\sum }_{n\ge 1}\mu (n)n^{-s})\left({\sum }_{n\ge 1}{n}^{1-s}\right)\mathrm{.}$

The first factor is $1\mathrm{/}\zeta (s)$ by [L3], and the second is $\zeta (s-1)$ by definition. Hence the product is $\zeta (s-1)\mathrm{/}\zeta (s)$.