KindiakMath

The Zeta Primes

For this post, we adopt the convention \mathbb N \equiv \mathbb N^+, so that 0 \notin \mathbb N.

Definition 1. Call a function f : \mathbb N \to \mathbb R multiplicative if for any m,n \in \mathbb N,

f(m \cdot n) = f(m) \cdot f(n).

Problem 1. Show that the map \displaystyle f(n) := 1/n^s is multiplicative.

(Click for Solution)

Solution. Using the laws of exponents,

\begin{aligned} f(m \cdot n) &= \frac 1{(m \cdot n)^s} \\ &= \frac 1{m^s \cdot n^s} \\ &= \frac 1{m^s} \cdot \frac 1{n^s} = f(m) \cdot f(n). \end{aligned}

Fix s > 1. Define the Riemann zeta function \zeta(s) by

\displaystyle \zeta(s) := \sum_{n \in \mathbb N} \frac 1{n^s} \equiv \sum_{n=1}^\infty \frac 1{n^s} = \frac 1{1^s} + \frac 1{2^s} + \frac 1{3^s} + \cdots.

Recall that the right-hand side converges by the p-series test.

Henceforth, let f be a non-negative multiplicative function such that

\displaystyle \sum_{n \in \mathbb N} f(n)

converges (absolutely).

Problem 2. Show that for any K \subseteq L \subseteq \mathbb N,

\displaystyle \sum_{n \in K} f(n) \leq \sum_{n \in L} f(n),

where both sums converge.

(Click for Solution)

Solution. Since K \subseteq L \subseteq \mathbb N, the indicator maps

\mathbb I_K(\cdot) : L \to \{0, 1\},\quad \mathbb I_L(\cdot) : \mathbb N \to \{0,1\}

are well-defined. By definition, since \mathbb I_K(\cdot), \mathbb I_L(\cdot) \in \{0, 1\},

\begin{aligned} 0 \leq \sum_{n \in K} f(n) &= \sum_{n \in L} f(n) \cdot \mathbb I_K(n) \\ &\leq \sum_{n \in L} f(n) \\ &= \sum_{n \in \mathbb N} f(n) \cdot \mathbb I_L(n) \leq \sum_{n \in \mathbb N} f(n). \end{aligned}

Since the right-hand side converges by hypothesis, by the comparison test, all sums on the left-hand side converge.

For any k \in \mathbb N, define the set

k^{\mathbb N} := \{ k^n : n \in \mathbb N\} = \{1, k, k^2, k^3, \dots\}.

Problem 3. For any k \in \mathbb N with k \geq 2, evaluate \displaystyle \sum_{n \in k^{\mathbb N}} \frac 1{n^s}.

(Click for Solution)

Solution. Since k \geq 2,

k^{-s} \leq 2^{-s} \leq 1/2 < 1.

By the convergent geometric series,

\begin{aligned} \sum_{n \in k^{\mathbb N}} \frac 1{n^s} &= \sum_{t = 0}^\infty \frac 1{(k^t)^s} = \sum_{t=0}^\infty (k^{-s})^t = \frac 1{1 - k^{-s}}. \end{aligned}

Now denote the set of prime numbers by \mathbf P \subseteq \mathbb N, which has been shown to be (countably) infinite. Enumerate them in ascending order:

\mathbf P = \{2, 3, 5, 7, 11, \dots\}

so that p_1 = 2, p_2 = 3, p_3 = 5, and so on.

For any pairs of non-empty sets K, L \subseteq \mathbb N, define

K \cdot L := \{xy : (x,y) \in K \times L\}.

We remark that (mn)^{\mathbb N} \subsetneq m^{\mathbb N} \cdot n^{\mathbb N}.

Problem 4. Show that for distinct p, q \in \mathbf P,

\displaystyle \left( \sum_{m \in p^{\mathbb N}} f(m) \right) \cdot \left( \sum_{n \in q^{\mathbb N}} f(n) \right) = \sum_{k \in p^{\mathbb N} \cdot q^{\mathbb N}} f(k).

(Click for Solution)

Solution. Thanks to absolute convergence, all sums here converge. Fix M, N \in \mathbb N. Then

\begin{aligned} \left( \sum_{i=0}^M f(p^i) \right) \cdot \left( \sum_{j=0}^N f(q^j) \right) &= \sum_{i=0}^M \left( f(p^i) \sum_{j=0}^N f(q^j) \right) \\ &= \sum_{i=0}^M \sum_{j=0}^N ( f(p^i) f(q^j) ) \\ &= \sum_{i=0}^M \sum_{j=0}^N f(p^i q^j). \end{aligned}

Then taking (M,N) \to \infty yields the desired result:

\begin{aligned} \left( \sum_{m \in p^{\mathbb N}} f(m) \right) \cdot \left( \sum_{n \in q^{\mathbb N}} f(n) \right) &= \left( \lim_{M \to \infty} \sum_{i=0}^M f(p^i) \right) \cdot \left( \lim_{N \to \infty} \sum_{j=0}^N f(q^j) \right) \\ &= \lim_{M \to \infty} \lim_{N \to \infty} \left( \sum_{i=0}^M f(p^i) \right) \cdot \left( \sum_{j=0}^N f(q^j) \right) \\ &= \lim_{(M,N) \to \infty} \sum_{i=0}^M \sum_{j=0}^N f(p^i q^j) \\ &= \sum_{k \in p^{\mathbb N} \cdot q^{\mathbb N}} f(k). \end{aligned}

For any n \in \mathbb N and x \in [1,\infty), define

\pi(n) := |\mathbf P \cap \{1,\dots, n\}|, \quad \pi(x) := \pi(\lfloor x \rfloor).

We call \pi(\cdot) : [1,\infty) \to \mathbb N the prime-counting function.

Inductively define

\displaystyle \prod_{i=1}^n K_i := \left( \prod_{i=1}^{n-1} K_i \right) \cdot K_n = K_1 \cdot K_2 \cdot \dots \cdot K_n.

Problem 5. Show that for any n \in \mathbb N,

\displaystyle \sum_{k=1}^n f(k) \leq \sum_{ \ell \in \prod_{i = 1}^{\pi(n)} p_i^{\mathbb N} } f(\ell) \leq \sum_{k \in \mathbb N} f(k).

In particular, evaluate

\displaystyle \sum_{ \ell \in \prod_{i = 1}^{\pi(n)} p_i^{\mathbb N} } f(\ell).

(Click for Solution)

Solution. By definition, \{1,\dots, n\} \subseteq \prod_{i=1}^{\pi(n)} p_i^{\mathbb N} \subseteq \mathbb N. The result follows from Problem 2. For the evaluation of the sum, use Problem 4:

\begin{aligned} \sum_{ \ell \in \prod_{i = 1}^{\pi(n)} p_i^{\mathbb N} } f(\ell) = \prod_{i=1}^{\pi(n)} \left( \sum_{j \in p_i^{\mathbb N}} f(j) \right). \end{aligned}

Problem 6. Explain why \pi(n) \to \infty as n \to \infty. Deduce the Euler product for the Riemann zeta function:

\displaystyle \zeta(s) = \prod_{p \in \mathbf P} \frac 1{1-p^{-s}}.

(Click for Solution)

Solution. We observe that for 2 \leq m \leq n, \mathbf P_m \subseteq \mathbf P_n implies

1 = \pi(2) \leq \pi(m) \leq \pi(n).

Furthermore, \pi(p_i) = i. Fix M > 0. Since \mathbf P is infinite, there exists at least M+1 primes p_1,\dots, p_{M+1}. Then for any n \geq p_{M+1},

\pi(n) \geq \pi(p_{M+1}) = M +1 > 0.

Hence, \pi(n) \to \infty. By Problem 5,

\displaystyle \sum_{k=1}^n f(k) \leq \prod_{i=1}^{\pi(n)} \left( \sum_{j \in p_i^{\mathbb N}} f(j) \right) \leq \sum_{k \in \mathbb N} f(k).

Taking n \to \infty on all sides,

\displaystyle \sum_{k=1}^{\infty} f(k) \leq \prod_{i=1}^{\infty} \left( \sum_{j \in p_i^{\mathbb N}} f(j) \right) \leq \sum_{k=1}^{\infty} f(k) .

Therefore,

\displaystyle \sum_{k=1}^{\infty} f(k) = \prod_{i=1}^{\infty} \left( \sum_{j \in p_i^{\mathbb N}} f(j) \right).

In the special case f(k) = 1/k^s, using Problem 3,

\begin{aligned} \zeta(s) = \sum_{k=1}^{\infty} \frac 1{k^s} &= \prod_{i=1}^{ \infty } \left( \sum_{j \in p_i^{\mathbb N}} \frac 1{j^s} \right) \\ &= \prod_{i=1}^{ \infty } \left( \frac 1{1 - p_i^{-s}} \right) = \prod_{ p \in \mathbf P } \left( \frac 1{1 - p^{-s}} \right). \end{aligned}

—Joel Kindiak, 25 Jan 26, 1624H

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