[proofplan]
We prove (ii) first, then derive (i). For (ii), unique factorisation gives $\prod_{p \leq x}(1 - 1/p)^{-1} \geq \sum_{n \leq x} 1/n$; the harmonic sum diverges, so the product diverges. For (i), we take logarithms of the product in (ii) and expand $-\log(1 - 1/p)$ as a power series. The first-order term is $\sum_{p \leq x} 1/p$ and the higher-order remainder is dominated by a convergent telescoping sum. Since the logarithm of the diverging product tends to infinity while the remainder is bounded, the prime harmonic sum must itself diverge.
[/proofplan]
[step:Expand the Euler product and compare with the harmonic sum]
Fix a real $x \geq 2$. Each factor $(1 - 1/p)^{-1}$ with $p$ prime equals the convergent geometric series
\begin{align*}
\left(1 - \frac{1}{p}\right)^{-1} = \sum_{k = 0}^{\infty} \frac{1}{p^k},
\end{align*}
since $0 < 1/p \leq 1/2 < 1$. All terms are positive, so we may multiply finitely many such series and rearrange without convergence issues:
\begin{align*}
\prod_{p \leq x} \left(1 - \frac{1}{p}\right)^{-1} = \prod_{p \leq x} \sum_{k_p = 0}^{\infty} \frac{1}{p^{k_p}} = \sum_{(k_p)_{p \leq x}} \frac{1}{\prod_{p \leq x} p^{k_p}},
\end{align*}
where the outer sum ranges over all tuples $(k_p)_{p \leq x}$ of non-negative integers indexed by the finitely many primes $p \leq x$. By the [Fundamental Theorem of Arithmetic](/theorems/???), each positive integer $n$ whose prime factors are all $\leq x$ corresponds to exactly one such tuple (via $n = \prod_{p \leq x} p^{k_p}$ with $k_p = v_p(n)$), and vice versa. Hence
\begin{align*}
\prod_{p \leq x} \left(1 - \frac{1}{p}\right)^{-1} = \sum_{\substack{n \geq 1 \\ p \mid n \Rightarrow p \leq x}} \frac{1}{n}.
\end{align*}
Now every integer $n \leq x$ has all its prime factors $\leq n \leq x$, so $\{n \geq 1 : n \leq x\} \subseteq \{n \geq 1 : p \mid n \Rightarrow p \leq x\}$. Since all terms $1/n$ are non-negative, we may restrict the sum to this subset — a strict reduction of the summation set — to obtain the lower bound
\begin{align*}
\prod_{p \leq x} \left(1 - \frac{1}{p}\right)^{-1} \geq \sum_{n \leq x} \frac{1}{n}.
\end{align*}
[/step]
[step:Conclude (ii) via divergence of the harmonic sum]
The harmonic sum satisfies $\sum_{n \leq x} 1/n \geq \int_1^{\lfloor x \rfloor + 1} d\mathcal{L}^1(t)/t = \log(\lfloor x \rfloor + 1) \to \infty$ as $x \to \infty$, where the inequality is the left Riemann sum comparison for the decreasing integrand $1/t$ on $[1, \lfloor x \rfloor + 1]$. Combining with Step 1,
\begin{align*}
\prod_{p \leq x} \left(1 - \frac{1}{p}\right)^{-1} \geq \sum_{n \leq x} \frac{1}{n} \to \infty \qquad \text{as } x \to \infty.
\end{align*}
This proves (ii).
[/step]
[step:Take logarithms and expand $-\log(1 - 1/p)$ to isolate $\sum 1/p$]
Since all factors $(1 - 1/p)^{-1} > 0$ are real and positive, and the product is finite for each fixed $x$, we may take the real natural logarithm:
\begin{align*}
\log \prod_{p \leq x} \left(1 - \frac{1}{p}\right)^{-1} = \sum_{p \leq x} \left[- \log\!\left(1 - \frac{1}{p}\right)\right].
\end{align*}
For each prime $p \geq 2$, $|1/p| \leq 1/2 < 1$, so the Taylor series
\begin{align*}
-\log(1 - u) = \sum_{k = 1}^{\infty} \frac{u^k}{k}, \qquad |u| < 1,
\end{align*}
converges at $u = 1/p$. Substituting,
\begin{align*}
-\log\!\left(1 - \frac{1}{p}\right) = \sum_{k = 1}^{\infty} \frac{1}{k p^k} = \frac{1}{p} + \sum_{k = 2}^{\infty} \frac{1}{k p^k}.
\end{align*}
Summing over primes $p \leq x$ (a finite sum) and rearranging the non-negative terms,
\begin{align*}
\log \prod_{p \leq x} \left(1 - \frac{1}{p}\right)^{-1} = \sum_{p \leq x} \frac{1}{p} + R_x,
\end{align*}
where we have defined the remainder
\begin{align*}
R_x := \sum_{p \leq x} \sum_{k = 2}^{\infty} \frac{1}{k p^k}.
\end{align*}
[/step]
[step:Bound the remainder $R_x$ uniformly in $x$]
Since $k \geq 2$ in the inner sum, $1/k \leq 1/2 \leq 1$, and we drop the factor $1/k$ to obtain the crude upper bound
\begin{align*}
R_x \leq \sum_{p \leq x} \sum_{k = 2}^{\infty} \frac{1}{p^k} = \sum_{p \leq x} \frac{p^{-2}}{1 - p^{-1}} = \sum_{p \leq x} \frac{1}{p(p - 1)},
\end{align*}
where we summed the geometric series with ratio $1/p \in (0, 1/2]$.
The sum over all integers $m \geq 2$ of $1/(m(m-1))$ telescopes:
\begin{align*}
\sum_{m = 2}^{\infty} \frac{1}{m(m - 1)} = \sum_{m = 2}^{\infty} \left( \frac{1}{m - 1} - \frac{1}{m} \right) = 1.
\end{align*}
Since every prime $p \leq x$ satisfies $p \geq 2$, and enlarging the index set from primes to all integers $m \geq 2$ only adds non-negative terms,
\begin{align*}
\sum_{p \leq x} \frac{1}{p(p - 1)} \leq \sum_{m = 2}^{\infty} \frac{1}{m(m - 1)} = 1.
\end{align*}
Hence $R_x \leq 1$ for every $x \geq 2$.
[/step]
[step:Conclude (i) by subtracting the bounded remainder from the divergent logarithm]
Combining Steps 3 and 4,
\begin{align*}
\sum_{p \leq x} \frac{1}{p} = \log \prod_{p \leq x} \left(1 - \frac{1}{p}\right)^{-1} - R_x.
\end{align*}
By (ii) proved in Step 2, the product on the right tends to $\infty$ as $x \to \infty$, so its logarithm does as well (since $\log$ is continuous and $\log t \to \infty$ as $t \to \infty$). By Step 4, $R_x \leq 1$ uniformly in $x$. Subtracting a bounded quantity from a sequence tending to $\infty$ still gives a sequence tending to $\infty$:
\begin{align*}
\sum_{p \leq x} \frac{1}{p} \geq \log \prod_{p \leq x} \left(1 - \frac{1}{p}\right)^{-1} - 1 \to \infty \qquad \text{as } x \to \infty.
\end{align*}
This establishes (i) and completes the proof.
[/step]
