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absolute convergence of a Dirichlet series depends only on the real part of its variable. The set of real parts giving convergence is upward closed by the comparison test. If that set is nonempty and not all of the real line, its finite infimum is the abscissa: every larger real part gives convergence, and every smaller real part gives divergence. Boundary convergence is a separate question.

For a cusp form $f=\sum c_n{q}^n$ and coprime integers $a,N$, the additive twist inserts the additive character $e^{2\pi ian/N}$. It is the Dirichlet series associated with the rational translate $f(a/N+\tau )$. Its completed Mellin transform uses the scale $N^s$ and the Gamma function.

Analytic number theory studies integers, prime numbers, and arithmetic functions using tools from real analysis and complex analysis, especially Dirichlet series and their singularities.

A smoothed form of the summatory totient function satisfiesUse the Dirichlet series $\zeta (s-1)/\zeta (s)$ and the kernel $x^s/[s(s+1)]$ to write the left side as a vertical contour integral. Move the line from real part greater than two to $3/2$. The only crossed pole is at two, with residue $x^3/(6\zeta (2))=x^3/{\pi }^2$; the remaining vertical integral is $O(x^{5/2})$ by the standard zeta bounds in the half-strip. If $A(x)={\sum }_{n\le x}\phi (n)$ and $S(x)={\int }_0^{x}A(t)dt$, positivity of the summands makes $A$ nondecreasing. The backward and forward difference quotients of $S$ bound $A(x)$. Insert $S(x)\sim x^3/{\pi }^2$, then let the relative difference-quotient step tend to zero, proving the displayed asymptotic. This monotone integral Tauberian lemma justifies the passage from a smoothed estimate to an unsmoothed one.

For $\Re s>0$, the alternating Dirichlet seriesconverges locally uniformly and defines a holomorphic function. For $\Re s>1$,

A Dirichlet polynomial is a finite Dirichlet series, $A(s)={\sum }_{n\le X}{a}_n{n}^{-s}$. Vertical integration separates equal product indices from oscillatory unequal ones.

For a positive integer-valued random variable of finite support, this transform is a finite Dirichlet series. Its values encode ordinary moments, $q_X(-r)=\mathbb{E}[X^r]$, and its derivatives at zero encode logarithmic moments: $(-1{)}^r{q}_X^{(r)}(0)=\mathbb{E}[(\log X{)}^r]$. Positive support is essential to these definitions; the transform is distinct from the ordinary probability generating function $\mathbb{E}[s^X]$.

A Dirichlet series is a series of the formMultiplication of absolutely convergent Dirichlet series corresponds to Dirichlet convolution of their coefficients.

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,

A generating function packages a sequence into a series so that operations on objects become operations on their encodings. An ordinary generating function uses the power series $A(z)={\sum }_{n\ge 0}{a}_n{z}^n$; in counting problems, the coefficient $a_n$ records the number of objects of size $n$. The variable $z$ initially marks size: it need not be assigned a numerical value. Coefficient extraction writes this recovery as $[z^n]A(z)=a_n$. One compact expression can therefore answer infinitely many counting questions, while explaining how their answers fit together.

Start with two contributions: choose $i$ from $\{0,1,2\}$ and $j$ from $\{0,1\}$, with exactly one option at each allowed value. Their individual ordinary generating functions are the polynomials $P(x)=1+x+x^2$ and $Q(x)=1+x$. To count ordered pairs by total $i+j$, multiply:The coefficients $1,2,2,1$ count the ways to obtain totals zero through three. For total two, the possibilities are $(1,1)$ and $(2,0)$; the two corresponding monomials both contribute $x^2$. There are six ordered pairs altogether, agreeing with $P(1)Q(1)=3\cdot 2=6$.

If the first contribution has two distinguishable choices when $i=1$, replace $P(x)$ by $1+2x+x^2$. Its product with $Q(x)$ becomes $1+3x+3x^2+x^3$. The coefficients now carry multiplicities as well as possible totals; the exponent still records cost. This lets the same construction handle several actions with identical costs without losing their separate identities.

In general, the Cauchy product givesFor each split of the total size, the rule of product counts the choices of the two components; summing over those splits counts all possibilities. This requires each combined object to have a unique decomposition into the two components. The same idea works for costs, lengths, or other additive measures, not just the number of pieces. See MIT 18.310, generating functions, sections 1–2.

Now cover a board of length $n$ with tiles of lengths one and two, without gaps or overlaps. Let $t_n$ count these tilings. There is exactly one empty tiling, so $t_0=1$, and $t_1=1$. Removing the first tile leaves a board shorter by one or two, giving the recurrence relationThese are the Fibonacci numbers shifted by one index: $t_n=F_{n+1}$ under the convention $F_0=0$, $F_1=1$. Keeping the empty tiling is essential to both the counting rule and the resulting ordinary generating function.

For $F(z)={\sum }_{n\ge 0}{t}_n{z}^n$, the empty board or a first tile followed by a remaining tiling givesEquivalently, $F(z)={\sum }_{m\ge 0}(z+z^2{)}^m$: each factor chooses one tile, and $m$ counts pieces while the exponent of $z$ records total length. The same ordinary generating function counts ordered histories of two repeatable actions costing one and two units. Histories $(1,2)$ and $(2,1)$ differ despite having the same total cost. MIT's notes use dots and dashes with these costs, including the empty message. ^[ref]

These identities hold in formal power series algebra: each requested coefficient uses only finitely many contributions. The repeated-choice expansion is legitimate because $z+z^2$ has zero constant term; a zero-cost repeatable action would allow infinitely many histories of the same size. Numerical evaluation separately requires a convergent series. Here the radius of convergence is $(\sqrt{5}-1)/2$, the reciprocal of the golden ratio. At $z=1$, the rational function equals $-1$, but the counting series diverges: that value cannot represent its sum.

Other generating functions change the encoding: an exponential generating function divides by factorials, while a Dirichlet series uses $n^{-s}$. A counting generating function also differs from a probability generating function, whose coefficients must be nonnegative and sum to one.

The L-function of a Hecke character is initially the displayed Euler product over prime ideals away from its conductor. The equivalent Dirichlet series sums $\psi (\mathfrak{a})N{\mathfrak{a}}^{-s}$ over integral ideals prime to the conductor. For a CM elliptic curve with its endomorphisms defined over the base, its Hasse-Weil L-function factors into the two conjugate Hecke L-functions.

If a Dirichlet series with nonnegative coefficients has finite abscissa of convergence ${\sigma }_c$, then the function it defines has a singularity at the real point $s={\sigma }_c$. Consequently, if that function extends holomorphically across every real point, the original series converges for every real $s$.

An L-function is an arithmetic function given initially by a Dirichlet series and often an Euler product, with analytic continuation and a functional equation in important cases. Dirichlet L-functions and L-functions of cusp forms are basic examples; p-adic L-functions interpolate suitably modified special values.

For $f={\sum }_{n\ge 1}{a}_n{q}^n$, define $L(f,s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ in a right half-plane. Its completion $(2\pi {)}^{-s}\Gamma (s)L(f,s)$ is the Mellin transform of $f(iy)$. The modular inversion and exponential cusp decay continue this completion to an entire function and give its reflection $s\mapsto k-s$. A normalized Hecke eigenform also has an Euler product of a Hecke eigenform.

If the Dirichlet series $F(s)={\sum }_n{a}_n{n}^{-s}$ is absolutely convergent at $\sigma >0$, thenThe scalar kernel is $(2\pi {)}^{-1}{\int }_{\mathbb{R}}{e}^{v(\sigma +it)}(\sigma +it{)}^{-2}dt=v_{+}$. This follows by Fourier inversion applied to $u{e}^{-\sigma u}{1}_{u\ge 0}$, whose Fourier transform is $(\sigma +it{)}^{-2}$. Absolute convergence of the series and of $\int |\sigma +it{|}^{-2}dt$ justifies interchange. The factor $ds=idt$ is essential.

For every positive integer $n$, the divisor sum of the Möbius function isThe sum includes all positive integer divisors, but only squarefree divisors contribute. Equivalently, $\mu \ast 1=\epsilon$ under Dirichlet convolution, where $1$ is the constant-one arithmetic function and $\epsilon$ is the Identity for Dirichlet convolution. ^[ref]

Each squarefree divisor chooses a subset of the distinct prime factors of $n$, contributing $(-1{)}^r$ when that subset has $r$ elements. Expanding a product therefore givesFor $n>1$ at least one prime factor occurs, so the product is zero. At $n=1$ it is an empty product, equal to one. This is the same alternating cancellation that drives inclusion-exclusion.

For $30=2\cdot 3\cdot 5$, all eight positive integer divisors are squarefree integers, givingPair each subset not containing $2$ with the subset obtained by inserting $2$: $(1,2),(3,6),(5,10),(15,30)$. Each pair cancels. The same nonzero terms occur for $60$, because its squarefree divisors are still the integer divisors of $30$; its other integer divisors have Möbius function value zero.

The Möbius divisor-sum identity is the cancellation behind Möbius inversion and reciprocal Dirichlet series: aggregation over integer divisors has an exact inverse.

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 $\mathrm{Re}s>1$.

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 $n\le x$, write $F(n)=F(x)-{\int }_n^x{F}^{\prime }(t)dt$, multiply by $f(n)$ and sum. Interchanging a finite sum with the integral gives the Abel summation formulaApply it with $F(t)=t^{-s}$ and $S(t)={\sum }_{n\le t}{a}_n$. If $\sigma >\alpha$, the boundary term is $O(x^{\alpha -\sigma })\to 0$ and ${\int }_1^{\infty }|S(t)t^{-s-1}|dt$ converges. Therefore the Dirichlet series converges andChoose $C$ with $|S(t)|\le C{t}^{\alpha }$ for $t\ge 1$. For $\sigma >\alpha +\delta$,The Weierstrass M-test proves uniform convergence of $\sum I_m$ on this entire open half-plane. For fixed $s$, the same estimate with its actual $\sigma$ gives $I_m(s)=O(m^{\alpha -1-\sigma })$. Each $I_m$ is an entire function, since its integral has bounded positive $t$ range and can be differentiated in $s$ under an integrable bound on each compact set. Thus the locally uniform sum is holomorphic, and so is its product with $s$. This gives holomorphy from polynomial growth of arithmetic partial sums and proves holomorphy of $A$ for $\mathrm{Re}s>\alpha$. Multiplication by the unbounded factor $s$ does not assert uniform convergence of $A$ on a whole unbounded half-plane; local uniform convergence is what the holomorphy argument needs.

With $a_n=\mu (n)$ and the hypothetical bound, $A_{\mu }(s)$ is holomorphic for $\mathrm{Re}s>3/4$. On $\mathrm{Re}s>1$, the Euler product gives $\zeta (s)A_{\mu }(s)=1$. The identity theorem extends this as a meromorphic identity throughout the connected half-plane $\mathrm{Re}s>3/4$. Hence $\zeta$ has no zeros in that half-plane: at any zero, the product with the holomorphic $A_{\mu }$ would be zero. The pole at one is not a zero; $A_{\mu }$ has a compensating zero there.

For $\mathrm{Re}s>1$, the Euler product theorem isThe local series converge absolutely, and the product is absolutely convergent in the sense that the sum over primes of the absolute deviations of the local factors from one is finite. Indeed this sum is bounded by ${\sum }_{n\ge 2}|f(n)|n^{-\mathrm{Re}s}$. To prove the identity, expand the product over primes at most $y$. Absolute convergence permits all those local expansions, and multiplicativity identifies the coefficients with $f(n)$ for integers whose prime factors are at most $y$. Every omitted integer exceeds $y$, so the difference from the full Dirichlet series is bounded by its absolute tail beyond $y$, which tends to zero. The same tail bound on a closed right half-plane inside $\mathrm{Re}s>1$ proves local uniform convergence. With signed coefficients a local factor can be zero; the identity does not require claiming every such general Euler product is nonzero.

For $d(n^2)$, multiplicativity gives local coefficients $d(p^{2e})=2e+1$. With $z=p^{-s}$,The divisor bound from (1), applied with a sufficiently small exponent, proves absolute convergence when $\mathrm{Re}s>1$; divergence at one follows from $d(n^2)\ge 1$. Multiplying the local identities therefore givesFor $n\le x$, the inequality $n^{\delta }\le x^{\delta }$ gives $d(n^2)/n\le x^{\delta }d(n^2)/n^{1+\delta }$. Sum those nonnegative terms and enlarge to the whole infinite series to get the required comparison.

For real $1<s<2$, integral comparison gives $\zeta (s)\le 1+(s-1{)}^{-1}\le 2/(s-1)$, while $\zeta (2s)\ge 1$. Hence $\zeta (s{)}^3/\zeta (2s)=O((s-1{)}^{-3})$ uniformly in that interval. Choosing $\delta =1/\log x$ for $x\ge 3$ makes $0<\delta <1$ and $x^{\delta }=e$. Thus