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The Möbius function has $\mu (1)=1$, $\mu (n)=0$ when $n$ is divisible by a prime square, and $\mu (n)=(-1{)}^r$ when $n$ is a product of $r$ distinct primes. The Riemann zeta function is initially the Dirichlet series $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$ for $\sigma =\Re s>1$. Both this Dirichlet series and $\sum \mu (n)n^{-s}$ have absolute convergence, because $|\mu (n)|\le 1$ and $\sum n^{-\sigma }<\infty$.

For a finite set of primes $p\le y$, multiplication of absolutely convergent geometric series and unique prime factorization giveEvery omitted integer has a prime factor exceeding $y$, so it exceeds $y$. The difference from $\zeta (s)$ has absolute value at most ${\sum }_{n>y}{n}^{-\sigma }\to 0$. This proves the Euler product, with the product interpreted as the limit of its finite products:

For the reciprocal identity, absolute convergence justifies multiplying and regrouping the two Dirichlet series by their product index. Their Dirichlet convolution coefficient at $m$ isThe coefficient calculation uses only the squarefree integers dividing $m$. Consequently the reciprocal identity is an identity of absolutely convergent series:

Let $M(x)={\sum }_{n\le x}\mu (n)$ be the Mertens function. Partial summation givesFor any $\Re s>0.9$, choose a sufficiently small $ϵ>0$ with $0.9+ϵ<\Re s$. The assumed bound makes the boundary term tend to zero and makes the integral absolutely convergent. Moreover, on each compact set of this half-plane, one can choose the same $ϵ$, so the integral has locally uniform convergence. Henceis a holomorphic function on $\Re s>0.9$, agreeing with the reciprocal Dirichlet series from part (b) on $\Re s>1$.

The identity theorem applied to the holomorphic functions $(s-1)\zeta (s)F(s)$ and $s-1$ extends their equality throughout $\Re s>0.9$; the assumed simple pole makes the first function holomorphic at $1$. Thus $\zeta (s)F(s)=1$ away from $1$, proving a zero-free half-plane:

To obtain the strongest symmetric conclusion, use the following basic facts explicitly: the functional equation of the Riemann zeta function gives the entire function $\xi (s)=\frac{1}{2}s(s-1){\pi }^{-s/2}\Gamma (s/2)\zeta (s)$ with $\xi (s)=\xi (1-s)$, and the trivial zeros of the Riemann zeta function are the negative even integers. Every Nontrivial zero of the Riemann zeta function is therefore accompanied by its reflection $1-s$. A Nontrivial zero of the Riemann zeta function with real part below $0.1$ would reflect to one above $0.9$, which is impossible. Consequently all nontrivial zeros lie in the closed stripThe $ϵ$ in the hypothesis prevents this argument from excluding either boundary line; the trivial zeros of the Riemann zeta function remain present.

Extend each Dirichlet character periodically to all integers by setting $\chi (n)=0$ when $\gcd (n,N)>1$. Its Dirichlet L-function is initially the absolutely convergent Dirichlet seriesThe group $(\mathbb{Z}/N\mathbb{Z}{)}^{\times }$ has $\phi (N)$ elements. Part (i) gives, for every integer $n$, including nonunits,For a nonunit both sides are zero, because the fixed class $a$ is a unit. Interchanging the finite character sum and the absolutely convergent Dirichlet series therefore provesThis is Orthogonality of Dirichlet characters applied to a residue-class indicator. Any meromorphic continuations of the two sides agree by uniqueness of analytic continuation.

For $\sigma >1$, both relevant Dirichlet series converge absolutely, so their product may be rearranged:The inner sum is one when $n=1$ and zero otherwise by Möbius inversion. HenceThis is also the reciprocal of the absolutely convergent Euler product.

Let $\hat{G}$ be the group of all Dirichlet characters modulo $q$, where $G=(\mathbb{Z}/q\mathbb{Z}{)}^{\times }$, and considerFor $p\nmid q$, let $f_p$ be the order of $p$ in $G$. The values $\chi (p)$ run through the $f_p$th roots of unity, each $|G|/f_p$ times, so the local factor isFor $p\mid q$ the local factor is one. Thus the Dirichlet series for $F$ has nonnegative coefficients.

The principal-character factor has a simple pole at $s=1$, while every nonprincipal Dirichlet L-function is entire. If some nonprincipal $L(1,\chi )$ vanished, its zero would cancel that pole and make $F$ entire. The Landau theorem for a Dirichlet series with nonnegative coefficients would then force the Dirichlet series of $F$ to converge for every real $s$. This is impossible: its coefficient at $m^{|G|}$ is at least one for every $m$ coprime to $q$, as is clear from the local factors. Hence

Take both functions to be the constant arithmetic function $1$. Thenso the divisor function is multiplicative by part (a). For $s$ with $\mathrm{Re}s>1$, the relevant Dirichlet series are absolutely convergent, so their Cauchy product may be rearranged:Equivalently, Dirichlet convolution becomes multiplication of absolutely convergent Dirichlet series.

For $\Re s>1$, the absolutely convergent Dirichlet seriesdefines the Riemann zeta function. To continue it, use partial summation in Stieltjes form:Since $0\le \{x\}<1$, the final integral converges locally uniformly for $\Re s>0$ and is holomorphic there. The displayed expression is consequently a meromorphic function on that half-plane, with its only pole at $s=1$. Since $s/(s-1)$ has residue one there, so does $\zeta$. Agreement in $\Re s>1$ makes this continuation unique by the identity theorem.

For $\sigma >1$, put $F(s)=-{\zeta }^{\prime }(s)/\zeta (s)$. Its absolutely convergent Dirichlet series andgive the three-four-one zero-free-region argumentThe pole of $\zeta$ at one gives

Suppose $\rho =\beta +i\gamma$ is a zero with $\gamma \ge 4$ and $\beta$ close to one. Apply the supplied Local partial-fraction expansion of the Riemann zeta logarithmic derivative at $\sigma +i\gamma$. Every term has positive real part, so retaining the term belonging to $\rho$ givesAt $\sigma +2i\gamma$ the same expansion gives merely $\Re F(\sigma +2i\gamma )\le O(\log \gamma )$. The zero is included in the supplied disk whenever $1-\beta$ and $\sigma -1$ are sufficiently small. Hence

Set $L=\log \gamma$ and $\sigma =1+a/L$, where $a>0$ is a sufficiently small fixed constant. If $(1-\beta )L$ were smaller than a sufficiently small constant $c>0$, division by $L$ would givea contradiction. Conjugation handles negative $\gamma$. Reducing $c$ to absorb the bounded range proves the classical zero-free region of the Riemann zeta function

The Mellin transform of $F$ isBecause the support is a compact subset of $(0,\infty )$, the integral defines an entire function of $s$. The Mellin inversion formula says that, for every real $\sigma$ and every $y>0$,

Apply this with $y=n/X$ and initially $\sigma =2$. Absolute convergence of the Dirichlet series for the logarithmic derivative permits interchange of sum and integral, givingTruncate at height $T=(\log X{)}^A$, where $A$ is a sufficiently large fixed constant. The assumed bound on $\tilde{F}$ makes the discarded tails smaller than the required error. The classical zero-free region of the Riemann zeta function and the standard bound ${\zeta }^{\prime }(s)/\zeta (s)\ll (\log (|t|+3){)}^{O(1)}$ there allow the truncated contour to move toThe only singularity crossed is the simple pole of $-{\zeta }^{\prime }/\zeta$ at $s=1$, whose residue is $1$. Its contribution isOn the new contour, $|X^s|=X^{1-c/\log \log X}$; the zeta bounds, contour length, and exponential decay of $\tilde{F}$ absorb into a slight decrease of $c$. ThereforeThis is the smoothed prime number theorem from a zero-free region.

If $f$ is a multiplicative function with $|f(n)|\le 1$, then for $\Re s>1$ its Dirichlet series has the Euler productIndeed, 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,so absolute convergence permits rearrangement and passage to the limit over all primes. This proves the formula.

For $\Re s>1$, the Dirichlet series multiplication rule and $-{\zeta }^{\prime }/\zeta (s)={\sum }_{n\ge 1}\Lambda (n)n^{-s}$ giveApply an effective Perron formula on the line $\kappa =1+1/\log x$ and truncate atThe bound from part (c) controls the truncation error.

Use the classical zero-free region of the Riemann zeta functiontogether with ${\zeta }^{\prime }/\zeta (s)\ll {\log }^2(|\Im s|+3)$ there. Contour shifting moves the Perron contour to $\Re s=1-c_1/\log T$. The only crossed singularity is the double pole at $s=1$, whose residue is $x\log x+Ax$ by part (b). On the new contour,and the logarithmic-derivative bounds contribute only powers of $\log x$, which can be absorbed by reducing the positive constant in the exponential. The horizontal integrals and Perron truncation error are $O(x\exp (-c_2\sqrt{\log x}))$ as well. Therefore, for some $c>0$,

Perron's formula recovers a summatory arithmetic function from its Dirichlet series by the inverse Mellin integralwith the usual convergence and endpoint conventions.

For real $w>1$, the finite Rankin–Selberg integral $I(f,f,w)$ has a nonnegative integrand. Tonelli theorem unfolds it without an a priori bound on the Dirichlet series. Integration over $x$ gives ${\int }_0^1|f(x+iy){|}^{2}dx={\sum }_{n\ge 1}|a_n{|}^2{e}^{-4\pi ny}$. A second application of Tonelli theorem gives the finite sum in the displayed identity with $s=w+k-1$. Positivity, rather than analytic continuation alone, proves absolute convergence.

For two weight-$k$ level-one cusp forms, use the primitive real-analytic Eisenstein series and $d\mu =dxdy/y^2$. The Rankin–Selberg unfolding identity isInitially this follows by absolute convergence, unfolding to $0\le x<1$, and integrating the Fourier series in $x$. Exponential cusp decay and polynomial growth of the primitive real-analytic Eisenstein series make the integral holomorphic for $\mathrm{Re}w>1$, continuing the Dirichlet series to $\mathrm{Re}s>k$.