Zeta and the Distribution of Primes

This is a standard fact assumed from The Riemann Zeta Function, but we recall the short argument for completeness, using the technique of comparison with an integral. We have $|n^{-s}| = n^{-\sigma}$. For $\sigma \ge \sigma_0 > 1$,

using that $x^{-\sigma}$ is decreasing in $x$ for $\sigma>0$ so each term $n^{-\sigma}$ ($n\ge2$) is bounded by $\int_{n-1}^n x^{-\sigma_0}dx$. The bound is uniform in $\sigma \ge \sigma_0$, giving uniform absolute convergence on $\sigma \ge \sigma_0$ for every $\sigma_0 > 1$. $\blacksquare$

Proof of the Theorem.

Step 1 (Euler product, part (a)). Fix $\sigma_0 > 1$ and work with $s$ satisfying $\sigma = \operatorname{Re}(s) \ge \sigma_0$. For each prime $p$, since $|p^{-s}| = p^{-\sigma} \le p^{-\sigma_0} < 1$, the geometric series

converges absolutely. Let $P$ be a finite set of primes and consider the finite product

Expanding this finite product of absolutely convergent series (justified because each factor is an absolutely convergent series, and a finite product of absolutely convergent series may be multiplied out term by term and rearranged freely), we get

where $S(P)$ is the multiplicative semigroup of positive integers all of whose prime factors lie in $P$ (this uses the Fundamental Theorem of Arithmetic: each $n \in S(P)$ corresponds to a unique tuple of exponents $(k_p)_{p\in P}$). Now let $P = P_N$ be the set of primes $\le N$. Every integer $n \le N$ has all its prime factors $\le N$ (trivially, since a prime factor of $n$ cannot exceed $n$), so $\{1,\dots,N\} \subseteq S(P_N)$. Hence

because every term with $n \le N$ cancels, and all surviving terms in both series correspond to $n > N$ (this uses that all terms of both series are subsets of $\{n^{-s}: n\ge1\}$, and the difference telescopes to include only indices missing from one side, all of which exceed $N$). By the Lemma, the tail $\sum_{n>N} n^{-\sigma} \to 0$ as $N \to \infty$, uniformly for $\sigma \ge \sigma_0$. Therefore

i.e. the infinite product $\prod_p (1-p^{-s})^{-1}$ converges to $\zeta(s)$, uniformly for $\sigma \ge \sigma_0 > 1$. Absolute convergence of the product (in the sense that $\sum_p |p^{-s}|$ converges) follows again from the Lemma since $\sum_p p^{-\sigma} \le \sum_n n^{-\sigma} < \infty$. An absolutely convergent infinite product of nonzero terms converges to a nonzero value, and each factor $(1-p^{-s})^{-1} \ne 0$; hence $\zeta(s) \ne 0$ for $\sigma > 1$. This proves part (a).

Step 2 (Logarithm of the Euler product). Since $\sigma > 1$ and $|p^{-s}| < 1$, we may take the principal branch logarithm of each factor:

the standard Taylor series for $-\log(1-z)$ at $z = p^{-s}$, valid since $|z|<1$. Summing over primes $p$ (justified below) gives, formally,

To justify taking $\log$ of the infinite product and interchanging sum and log continuously, we use that $\log$ of a convergent product of nonzero terms (with the product absolutely convergent, as shown in Step 1) equals the sum of the logs of the factors, provided we track branches consistently; since each factor is within a fixed disk around $1$ (as $|p^{-s}|\le p^{-\sigma_0} < 1$ makes each factor close to $1$ for large $p$), the principal branch is consistent and the identity

holds as an absolutely convergent double sum, because

using $\sigma \ge \sigma_0$ and $p\ge2$. Absolute convergence of the double sum permits rearrangement into a single sum over $n = p^k$:

This uses unique factorization: each pair $(p,k)$ corresponds to a unique prime power $n=p^k$, so no double-counting occurs and every prime power arises exactly once.

Step 3 (Differentiate term by term). The series $\sum_n c(n) n^{-s}$ converges absolutely and, restricted to any half-plane $\sigma \ge \sigma_0>1$, converges uniformly (same estimate as above, uniform in $\sigma\ge\sigma_0$, and this is inherited by the derivative series below since $\log n \le n^{\varepsilon}$ for any $\varepsilon>0$ and large $n$, which preserves convergence after an arbitrarily small loss of exponent — concretely, for $\sigma > \sigma_1 > 1$ we can absorb $\log n$ into a slightly smaller exponent margin). A standard theorem on term-by-term differentiation of a locally uniformly convergent series of holomorphic functions (each term $c(n)n^{-s} = c(n)e^{-s\log n}$ is entire in $s$) states: if $\sum f_n(s)$ converges to $F(s)$ locally uniformly on an open set, then $F$ is holomorphic there and $F'(s) = \sum f_n'(s)$, the derivative series also converging locally uniformly. Applying this with $f_n(s) = c(n) n^{-s}$, so $f_n'(s) = -c(n)(\log n)\, n^{-s}$, we obtain, for $\sigma>1$,

(Here we also used $\frac{d}{ds}\log \zeta(s) = \zeta'(s)/\zeta(s)$, valid since $\zeta(s)\ne 0$ for $\sigma>1$ by Step 1, so $\log\zeta(s)$ is a genuine holomorphic branch there with the usual derivative rule.)

Step 4 (Identify the coefficients as $\Lambda(n)$). For $n = p^k$ we compute

and for $n$ not a prime power, $c(n) = 0 = \Lambda(n)$. Hence $c(n)\log n = \Lambda(n)$ for every $n \ge 2$, and trivially $\Lambda(1)=0$ matches $n=1$ contributing nothing. Substituting into Step 3:

Absolute convergence of this final series for $\sigma > 1$ follows since it is (up to sign) the derivative series established convergent in Step 3. This proves part (b) and completes the theorem. $\blacksquare$

Write $n = p_1^{a_1}\cdots p_r^{a_r}$ by unique factorization, so $\log n = \sum_{i=1}^r a_i \log p_i$. On the other hand, the divisors $d$ of $n$ with $\Lambda(d)\ne 0$ are exactly the prime-power divisors $p_i^{j}$ for $1\le j \le a_i$, $1\le i \le r$, each contributing $\Lambda(p_i^j) = \log p_i$. Summing,

proving the first identity for every $n\ge1$ (for $n=1$ both sides are $0$, the empty sum). For the second identity, apply the Möbius inversion formula: if $f(n) = \sum_{d\mid n} g(d)$ for all $n\ge1$, then $g(n) = \sum_{d\mid n}\mu(d) f(n/d)$. This formula is proved (assumed known from The Mobius Function) using $\sum_{d\mid m}\mu(d) = [m=1]$. Taking $f(n)=\log n$ and $g(n)=\Lambda(n)$, which satisfy $f=\sum_{d\mid n} g(d)$ by the first identity, inversion gives