The Euler Product

$\zeta(s) = \sum_{n\ge 1} n^{-s}$ 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.

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, $\prod_{p}\left(1 - \frac{1}{p^2}\right) = \frac{6}{\pi^2} \approx 0.6079271$.

This is interpreted as a natural density: the density of coprime pairs among integer pairs is $\frac{6}{\pi^2}$ — a density, not a probability on a probability space, since there is no uniform probability distribution on the integers.

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.

Justify taking logs for $s > 1$: for a finite partial product, $\log$ turns the product into a sum of logs; the partial products converge to $\zeta(s) > 0$, so continuity of $\log$ passes this to the limit.

Using $-\log(1 - x) = \sum_{j\ge 1} \frac{x^j}{j}$ (Taylor series, $|x| < 1$) gives $\log \zeta(s) = \sum_{p}\sum_{j\ge 1} \frac{1}{j\,p^{js}}$.

Every term $\frac{1}{j\,p^{js}}$ is positive, so Tonelli's theorem for counting measure permits any rearrangement of the double sum, and the double sum converges absolutely for $s > 1$.

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 $s \to 1^+$.

Set up the monotone convergence step: for each prime $p$, $p^{-s}$ increases to $p^{-1}$ as $s \downarrow 1$, and every term is nonnegative, so the monotone convergence theorem for counting measure gives $\sum_{p} p^{-s} \uparrow \sum_{p} p^{-1}$ as $s \to 1^+$ (the limit may be $+\infty$).

Hence if $\sum_{p} \frac{1}{p}$ converged to a finite value at $s = 1$, $\sum_{p} p^{-s}$ would have 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.