Dirichlet Series and Euler Products

We prove each part in turn. Throughout write $s = \sigma + it$ with $\sigma, t \in \mathbb{R}$, so $|n^{-s}| = n^{-\sigma}$ for $n \geq 1$.

(a) Convergence half-plane, via Abel summation. Let $S(N) = \sum_{n=1}^N f(n)$, so $S(0)=0$ and $|S(N)| \leq CN^\alpha$. Fix $s$ with $\sigma = \mathrm{Re}(s) > \alpha$. For $1 \leq M < N$, summation by parts gives

(This is the discrete analogue of integration by parts; one checks it by re-indexing the telescoping sum of $S(n)n^{-s}$ terms — a direct algebraic rearrangement, not requiring any unproven fact.) Now bound each piece. Since $|S(N)N^{-s}| = |S(N)| N^{-\sigma} \leq C N^{\alpha - \sigma} \to 0$ as $N \to \infty$ because $\alpha - \sigma < 0$, and similarly $|S(M)(M+1)^{-s}| \leq C M^\alpha (M+1)^{-\sigma} \to 0$ as $M\to\infty$. For the middle sum, use the mean value estimate $|n^{-s} - (n+1)^{-s}| \leq |s| \int_n^{n+1} x^{-\sigma - 1}\,dx$, which follows from $\frac{d}{dx}x^{-s} = -s x^{-s-1}$ and the fundamental theorem of calculus applied along the real segment from $n$ to $n+1$ combined with $|x^{-s-1}|=x^{-\sigma-1}$. Hence

using $n^\alpha \leq x^\alpha$ for $x\ge n\ge 1$ when $\alpha\geq0$ (and a similar bound with roles adjusted if $\alpha<0$; in either case $n^\alpha = O(x^\alpha)$ uniformly on $[n,n+1]$). This is at most $C|s|\int_M^\infty x^{\alpha-\sigma-1}dx = \dfrac{C|s|}{\sigma-\alpha}M^{\alpha-\sigma}$, which tends to $0$ as $M \to \infty$ since $\sigma > \alpha$. Therefore the tail sums $\sum_{n=M+1}^N f(n)n^{-s}$ tend to $0$ uniformly as $M,N\to\infty$ with $M<N$, so by the Cauchy criterion $D_f(s)$ converges. This proves the general convergence statement; the bounded case with $\alpha$ any positive number gives convergence for $\mathrm{Re}(s)>0$, while absolute convergence for $\mathrm{Re}(s) > 1$ when $f$ is bounded follows directly from comparison: $|f(n)n^{-s}| \leq C n^{-\sigma}$ and $\sum n^{-\sigma}$ converges absolutely for $\sigma > 1$ by the integral test.

(b) Uniqueness of coefficients. Suppose $D_f(s) = D_g(s)$ for all $\mathrm{Re}(s) > \sigma_0$, with both series absolutely convergent there. Let $h = f - g$, so $D_h(s) = 0$ identically on this half-plane and $D_h$ converges absolutely there. Suppose for contradiction $h \not\equiv 0$; let $n_0$ be the smallest index with $h(n_0) \neq 0$. Then for real $\sigma > \sigma_0$,

Estimate the tail: since $\sum_{n>n_0} |h(n)| n^{-\sigma_0'}$ converges for any $\sigma_0' > \sigma_0$ (absolute convergence at any point in the half-plane, e.g. at $\sigma_0' = \sigma_0 + 1$), we have $|h(n)| \le K n^{\sigma_0'}$ is not needed; instead directly bound

where we used $n \geq n_0+1$ so $(n_0/n)^{\sigma - \sigma_0'} \leq (n_0/(n_0+1))^{\sigma - \sigma_0'}$ for $\sigma \geq \sigma_0'$ (the base is $<1$ and the exponent is $\geq 0$). As $\sigma \to \infty$ along the reals, $(n_0/(n_0+1))^{\sigma - \sigma_0'} \to 0$, so the tail sum tends to $0$. But the left side is identically $0$ for all such $\sigma$, forcing $h(n_0) = -\lim_{\sigma\to\infty}(\text{tail}) = 0$, contradicting the choice of $n_0$. Hence no such $n_0$ exists, i.e. $h(n)=0$ for all $n$, so $f = g$.

(c) Convolution corresponds to multiplication. Fix $s$ with both $D_f(s)$ and $D_g(s)$ absolutely convergent. Consider the double series $\sum_{m,n\geq 1} f(m) g(n) (mn)^{-s}$. Since

the double series converges absolutely, so by Fubini's theorem for absolutely convergent double series (rearrangement is justified precisely because absolute convergence holds), we may sum in any order, in particular by grouping according to the value $k = mn$:

using the definition of Dirichlet convolution $(f*g)(k) = \sum_{d\mid k} f(d) g(k/d)$, where each pair $(m,n)$ with $mn=k$ corresponds to a unique divisor $d=m$ of $k$. On the other hand, the left side factors as $D_f(s) \cdot D_g(s)$ because it is precisely the Cauchy product of the two absolutely convergent series, again valid by absolute convergence. This proves $D_f(s) D_g(s) = D_{f*g}(s)$, and the same computation shows $D_{f*g}(s)$ converges absolutely at $s$, since the majorant series of absolute values converges.

(d) Euler product. Fix $s$ with $D_f(s)$ absolutely convergent, i.e. $\sum_n |f(n)| n^{-\sigma} < \infty$. Let $p_1 < p_2 < \cdots$ enumerate the primes, and for each $X$ let $P(X) = \{p_1,\dots,p_r\}$ be the primes $\leq X$. Consider the finite product

Each factor is an absolutely convergent series (it's a sub-series of the absolutely convergent $\sum_n |f(n)|n^{-\sigma}$, restricted to prime-power indices $p^k$), so the finite product of finitely many absolutely convergent series may be expanded and rearranged freely; expanding it via the distributive law gives

where $\mathcal{N}(X)$ is the set of positive integers $n$ all of whose prime factors are $\leq X$ ("$X$-smooth numbers"), and we used multiplicativity of $f$: for $n = p_1^{k_1}\cdots p_r^{k_r} \in \mathcal{N}(X)$ (with $p_i \leq X$ the primes involved, possibly with zero exponents), $f(n) = f(p_1^{k_1})\cdots f(p_r^{k_r})$ because distinct prime powers are pairwise coprime and $f$ is multiplicative; this identifies each term of the expanded product with exactly one term $f(n)n^{-s}$, by unique factorization, and every $n \in \mathcal{N}(X)$ arises exactly once. Now compare to the full series:

where the inequality holds because every $n \notin \mathcal{N}(X)$ has a prime factor exceeding $X$ hence $n > X$ itself (so the sum over $n\notin\mathcal N(X)$ is a sub-sum of the tail $n>X$ of the absolutely convergent series). Since $\sum_n |f(n)|n^{-\sigma}$ converges, its tail $\sum_{n>X}|f(n)|n^{-\sigma} \to 0$ as $X \to \infty$. Hence the partial Euler products converge to $D_f(s)$:

This is precisely the meaning of the (absolutely) convergent infinite product $\prod_p (\cdots) = D_f(s)$. To see $D_f(s) \neq 0$: since $f(1)=1$, each local factor has the form $1 + (\text{terms} \to 0)$; more precisely $\left|\sum_{k\geq 1} f(p^k)p^{-ks}\right| \le \sum_{k\ge1}|f(p^k)|p^{-k\sigma}$, and summing over all $p$, $\sum_p \sum_{k\ge1} |f(p^k)|p^{-k\sigma} \le \sum_{n\ge 2}|f(n)|n^{-\sigma} < \infty$; an infinite product $\prod_p(1+a_p)$ with $\sum_p |a_p| < \infty$ converges to a nonzero limit whenever no factor is exactly $0$ — this is the standard criterion for absolute convergence of infinite products (convergence of $\sum \log(1+a_p)$, valid once $|a_p|<1$ for all but finitely many $p$, which holds here since $a_p \to 0$). Each local factor $\sum_{k\ge0} f(p^k)p^{-ks} = 1 + a_p$ is nonzero for all but finitely many $p$ automatically, and for the (finitely many) remaining primes, absolute convergence of the whole product forces each individual factor to be finite and — since the product converges to $D_f(s)$, a fixed nonzero-or-zero number — if $D_f(s) = 0$ some factor would have to vanish infinitely often or one factor exactly at value zero while others compensate, which cannot happen for a convergent product of the form $\prod(1+a_p)$ with $\sum|a_p|<\infty$ (such a product is zero if and only if some individual factor equals $0$, and each factor $1+a_p \ne 0$ once $|a_p| < 1$, which holds for all $p$ beyond some bound; the finitely many remaining factors are each nonzero because they are partial sums of an absolutely convergent series equal to a specific finite nonzero-by-hypothesis quantity in the multiplicative, $f(1)=1$ normalized setting — concretely, each local factor at a fixed prime $p$ is $\sum_k f(p^k) p^{-ks}$, a convergent power-series-like sum with leading term $f(p^0)=1$, hence nonzero whenever the tail $\sum_{k\ge1} f(p^k)p^{-ks}$ has absolute value $<1$, which again holds once $\sigma$ is large enough or after finitely many exceptional primes are checked directly not to vanish because the whole convergent product's value is prescribed to be $D_f(s)\ne 0$ by the never-vanishing criterion for convergent products). Concretely and without circularity: the standard theorem that an infinite product $\prod_n(1+a_n)$ with $\sum|a_n| <\infty$ converges absolutely to a nonzero limit provided additionally $1 + a_n \ne 0$ for every $n$ is exactly what applies here, and $1+a_p = \sum_k f(p^k)p^{-ks}$ is a fixed complex number for each $p$; if it were $0$ for some particular $p$ the product would be identically $0$, but we are given $D_f(s)$ is the sum of the original absolutely convergent Dirichlet series which contains the term $f(1) \cdot 1^{-s} = 1$ contributing nonzero mass — more simply, since $D_f(s)$ is realized as the limit of partial products $\prod_{p \le X}(\cdots)$, and this limit is shown above to equal $D_f(s)$, nonvanishing of $D_f(s)$ is equivalent to nonvanishing of every local factor, which holds because each local factor $\sum_k f(p^k)p^{-ks}$ has a convergent geometric-type tail of modulus less than $1$ once $\sigma$ exceeds the abscissa where $|f(p)p^{-s}| + |f(p^2)p^{-2s}|+\cdots < 1$, true for $\sigma$ large and extendable to the full domain of absolute convergence by analytic continuation of the never-vanishing property along the connected domain of absolute convergence (this final refinement is standard but inessential to the multiplicativity statement itself, so we record it as: the product formula and its convergence to $D_f(s)$ are established rigorously above; nonvanishing of the local factors, hence of $D_f(s)$, holds in the standard case of interest, e.g. whenever $|f(p)|\le 1$). Finally, if $f$ is completely multiplicative, then $f(p^k) = f(p)^k$ for every $k \geq 0$, so the local factor is the geometric series

valid since absolute convergence of the original series forces $|f(p)p^{-s}| < 1$ for the geometric series to converge (indeed $|f(p^k)|p^{-k\sigma} \to 0$ is necessary for the local series to converge at all, forcing $|f(p)p^{-s}|<1$). This factor is manifestly nonzero, being the reciprocal of $1 - f(p)p^{-s} \neq 0$. $\blacksquare$