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  "body": "\n## Proof\n\nWe argue by contradiction and count the integers up to $x$ in two ways.\n\n1. By [[Euclid's theorem|0292EPPPT0]], there are infinitely many [[prime numbers|0SN0MAEKYX]]. List them in increasing order as $p_1 \u003c p_2 \u003c p_3 \u003c \\cdots$.\n2. For $j \\ge 1$, put $s_j = \\sum_{i=1}^{j} 1/p_i$. Every partial sum $\\sum_{p \\le x} 1/p$ equals some $s_j$ or $0$, and every $s_j$ is such a partial sum.\n3. Suppose the partial sums are bounded. Then the set $\\set{s_j : j \\ge 1}$ is bounded above. By the [[completeness axiom|0AXWET80YW]], it has a supremum $L$.\n4. By the [[approximation property of the supremum|0M8TGGP9JP]], there is an index $k$ with $s_k \u003e L - 1/2$.\n5. For every $j \u003e k$, we get $\\sum_{i=k+1}^{j} 1/p_i = s_j - s_k \u003c L - (L - 1/2) = 1/2$, because $s_j \\le L$.\n6. Put $x = 2^{2k+2}$. Let $M$ be the set of integers $n$ with $1 \\le n \\le x$ such that no prime $p_i$ with $i \u003e k$ [[divides|096PW7Q014]] $n$.\n7. Upper estimate. Let $n \\in M$. By [[the square times square-free decomposition|080Q3RPKY6]], $n = m^2 r$ with positive integers $m$, $r$ and $r$ [[square-free|004JW7A2N5]].\n8. Every prime that divides $r$ divides $n$, so it is one of $p_1, \\dots, p_k$.\n9. By the [[existence of prime factorization|020K0X7HVR]], $r$ is a product of primes. No prime occurs twice, since then its square would divide $r$.\n10. So $r$ is the product of a subset of $\\set{p_1, \\dots, p_k}$. There are $2^k$ subsets, so $r$ takes at most $2^k$ values.\n11. Also $m^2 \\le n \\le x$, so $m \\le \\sqrt{x} = 2^{k+1}$. So $m$ takes at most $2^{k+1}$ values.\n12. The pair $(m, r)$ determines $n$. Hence $|M| \\le 2^k \\cdot 2^{k+1} = 2^{2k+1} = x/2$.\n13. Lower estimate. Let $1 \\le n \\le x$ with $n \\notin M$. Then some $p_i$ with $i \u003e k$ divides $n$, and $p_i \\le n \\le x$.\n14. Let $J$ be the largest index with $p_J \\le x$. If $J \\le k$, no such $n$ exists, and $|M| = x \u003e x/2$.\n15. If $J \u003e k$, each such $n$ is a multiple of some $p_i$ with $k \u003c i \\le J$. The number of multiples of $p_i$ in $\\set{1, \\dots, x}$ is at most $x / p_i$.\n16. So $x - |M| \\le \\sum_{i=k+1}^{J} x / p_i = x \\sum_{i=k+1}^{J} 1/p_i \u003c x/2$ by step 5. Hence $|M| \u003e x/2$.\n17. Steps 12, 14 and 16 give $x/2 \u003c |M| \\le x/2$, which is a contradiction.\n18. So the partial sums $\\sum_{p \\le x} 1/p$ are not bounded, and the series $\\sum_p 1/p$ diverges.\n\n## Idea\n\nIf the series converged, the primes beyond some $p_k$ would have reciprocal sum below $1/2$. Then fewer than half of the numbers up to $x$ could have a large prime factor. But numbers built only from $p_1, \\dots, p_k$ are rare: writing them as a square times a square-free part shows there are at most $2^k \\sqrt{x}$ of them. For $x = 4^{k+1}$ the two counts cannot both hold.\n",
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