The Euler Product

Prove the Euler product $\zeta(s) = \prod_{p\ \text{prime}} \frac{1}{1 - p^{-s}}$ for $s > 1$ by a sieve argument. Hint: $\zeta(s)(1 - 2^{-s})$ removes all even terms.

\zeta(s)(1 - 2^{-s}) = \sum_{n \ge 1} \frac{1}{n^s} - \sum_{n \ge 1} \frac{1}{(2n)^s} = \sum_{2 \nmid n} \frac{1}{n^s}

\zeta(s)\prod_{i=1}^{k}(1 - p_i^{-s}) = 1 + \sum_{\substack{n > 1 \\ (n, P_k) = 1}} \frac{1}{n^s}, \qquad P_k := \prod_{i=1}^{k} p_i

k = 6

(sieved through prime

13

):

P_6 \approx 1.6179177

, error

\zeta(2) - P_6 \approx 0.0270164

— each extra sieve layer pushes the product closer to

\zeta(2)

.

Show proof

$\zeta(s) = \sum_{n\ge 1} n^{-s}$ converges absolutely for $s > 1$ , so its terms may be freely rearranged and regrouped. Since $\frac{1}{1 - 2^{-s}} = \sum_{j\ge 0} 2^{-js}$ (a geometric series with ratio $2^{-s} < 1$ ), multiplying gives $\sum_{n} n^{-s} - \sum_{n} (2n)^{-s}$ : every even term cancels, leaving only the odd terms: $\zeta(s)(1 - 2^{-s}) = \sum_{2\nmid n} n^{-s}$ .
Do the same for the next prime $3$ : $(1 - 3^{-s})\sum_{2\nmid n} n^{-s} = \sum_{2\nmid n} n^{-s} - \sum_{2\nmid n} (3n)^{-s}$ , killing every multiple of $3$ . Repeating for $p_1 = 2, p_2 = 3, \dots, p_k$ gives $\zeta(s)\prod_{i=1}^{k}(1 - p_i^{-s}) = 1 + \sum_{n > 1,\ (n,P_k)=1} n^{-s}$ ( $P_k = p_1 \cdots p_k$ ): only the terms whose $n$ shares no factor with the first $k$ primes survive.
The leftover tail is squeezed to zero: $0 \le \sum_{n > 1,\ (n,P_k)=1} n^{-s} \le \sum_{n > p_k} n^{-s} \xrightarrow[k\to\infty]{} 0$ , because any $n > 1$ coprime to $P_k$ has all prime factors $\ge p_{k+1} > p_k$ , hence $n \ge p_{k+1} > p_k$ . So $\zeta(s)\prod_{i=1}^{k}(1 - p_i^{-s}) \to 1$ , and inverting gives $\zeta(s) = \prod_{p}\frac{1}{1 - p^{-s}}$ for $s > 1$ .
Numerical check ( $s = 2$ , $k = 6$ ): $P_6 = \prod_{i=1}^{6}\frac{1}{1-p_i^{-2}} \approx 1.6179177$ with error $\zeta(2) - P_6 \approx 1.6449341 - 1.6179177 = 0.0270164$ , matching the caption — each sieve layer peels off one more layer of the tail sum.

Use the Basel result $\zeta(2) = \frac{\pi^2}{6}$ to show $\prod_{p\ \text{prime}} \left(1 - \frac{1}{p^2}\right)^{-1} = \frac{\pi^2}{6}$ .

\prod_{p}\left(1 - \frac{1}{p^2}\right)^{-1} = \zeta(2) = \frac{\pi^2}{6}

P_3 = \frac{1}{1 - \frac14}\cdot\frac{1}{1 - \frac19}\cdot\frac{1}{1 - \frac{1}{25}} = \frac{4}{3}\cdot\frac{9}{8}\cdot\frac{25}{24} = \frac{25}{16} = 1.5625

k = 3

(primes

2, 3, 5

):

P_3 = \frac{25}{16} = 1.5625

, error

\frac{\pi^2}{6} - P_3 \approx 0.0824341

— still about

5\%

short, but the increase is monotone.

Show proof

Problem 1's Euler product holds for every $s > 1$ , so substituting $s = 2$ gives $\zeta(2) = \prod_{p}\frac{1}{1 - p^{-2}} = \prod_{p}\left(1 - \frac{1}{p^2}\right)^{-1}$ directly: the sum side and the product side point at the same limit.
The Basel result (see the Basel scene) is $\zeta(2) = \sum_{n\ge 1} \frac{1}{n^2} = \frac{\pi^2}{6}$ . Hence the product over primes $\prod_{p}\left(1 - \frac{1}{p^2}\right)^{-1}$ also converges to $\frac{\pi^2}{6} \approx 1.6449341$ .
Each factor $\frac{1}{1 - p^{-2}} > 1$ , so the partial products $P_k$ increase monotonically and are bounded above by $\zeta(2)$ , hence converge. Inverted, $\prod_{p}\left(1 - \frac{1}{p^2}\right) = \frac{6}{\pi^2} \approx 0.6079271$ — also the probability that two random integers are coprime.
Numerical check ( $k = 3$ ): with just the first three primes $2, 3, 5$ , $P_3 = \frac{4}{3}\cdot\frac{9}{8}\cdot\frac{25}{24} = \frac{25}{16} = 1.5625$ , and the error is $\frac{\pi^2}{6} - \frac{25}{16} \approx 1.6449341 - 1.5625 = 0.0824341$ , matching the caption.

Prove that the sum of prime reciprocals $\sum_{p\ \text{prime}} \frac{1}{p}$ diverges, starting from the Euler product as $s \to 1^+$ .

\log \zeta(s) = \sum_{p}\sum_{j\ge 1} \frac{1}{j\,p^{js}} = \sum_{p} \frac{1}{p^s} + O(1) \qquad (s > 1)

\zeta(s) \ge \sum_{n \le 1/(s-1)} \frac{1}{n^s} \ge \frac{1}{s - 1} \;\Rightarrow\; \log \zeta(s) \to +\infty \quad (s \to 1^+)

k = 15

(through prime

47

):

P_{15} \approx 1.6385679631

, error

\approx 0.0063661

— the product converges, but as

s \to 1^+

the log of the same product diverges, dragging the prime-reciprocal sum up with it.

Show proof

Taking logs of the Euler product for $s > 1$ and using $-\log(1 - x) = \sum_{j\ge 1} \frac{x^j}{j}$ (Taylor series, $|x| < 1$ ) gives $\log \zeta(s) = \sum_{p}\sum_{j\ge 1} \frac{1}{j\,p^{js}}$ . Splitting off $j = 1$ yields $\log \zeta(s) = \sum_{p} \frac{1}{p^s} + \sum_{p}\sum_{j\ge 2} \frac{1}{j\,p^{js}}$ .
The $j \ge 2$ double sum is uniformly bounded for $s \ge 1$ : $\sum_{p}\sum_{j\ge 2} \frac{1}{j\,p^{js}} \le \sum_{p} \frac{1}{p^2}\sum_{j\ge 2} 2^{1-j} = \sum_{p} \frac{1}{p^2} < \sum_{n\ge 2} \frac{1}{n^2} = \zeta(2) - 1 = \frac{\pi^2}{6} - 1 \approx 0.6449341$ , so that part is $O(1)$ . Hence $\log \zeta(s) = \sum_{p} \frac{1}{p^s} + O(1)$ for $s > 1$ .
Meanwhile $\zeta(s) \ge \int_1^{\infty} x^{-s}\,dx = \frac{1}{s-1}$ — just sum $n^{-s} \ge \int_n^{n+1} x^{-s}\,dx$ over $n$ . Hence $\log \zeta(s) \ge -\log(s - 1) \to +\infty$ as $s \to 1^+$ .
Since $\log \zeta(s) = \sum_{p} p^{-s} + O(1) \to +\infty$ , we get $\sum_{p} p^{-s} \to +\infty$ as $s \to 1^+$ . If $\sum_{p} \frac{1}{p}$ converged at $s = 1$ , the monotone convergence theorem would force $\sum_{p} p^{-s}$ to approach that sum as $s \to 1^+$ — a contradiction. Therefore $\sum_{p} \frac{1}{p} = +\infty$ : the sum of prime reciprocals diverges.
Numerical feel: at $s = 2$ , $P_{15} \approx 1.6385679631$ sits almost on $\zeta(2) \approx 1.6449341$ (error $\approx 0.0063661$ , matching the caption), but as $s$ slides toward $1$ , $\log \zeta(s)$ blows up to infinity, and with the $O(1)$ part accounted for, only $\sum_p p^{-s}$ can supply that infinity.

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