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Prove that the infinite sum $\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{5}$ + $\frac{1}{7}$... diverges.

Here is an attempt to prove this by contradiction.

Let P denote the set of all prime number, then we want to show $\sum_ {p \in P} \frac{1}{p}$ diverges.

Suppose $\sum_ {p \in P} \frac{1}{p} \to l \ (converges)$ then $\exists \ m \in \mathbb{N} \ s.t. \sum_ {p \in P \ p \ \le \ m} \frac{1}{p} \ge \ l-\frac{1}{2} $

let $P_{small} = \{p \in P, p>m \}$, and let its size $|P_{small}| = k,$ and write $P_{small} = \{P_1, P_2, P_3,...P_k\}$

and similarly let $P_{large} = \{ p\in P, p>m\} $

then we get $\sum_ {p \in P_{small}} \frac{1}{p} \ge \ l-\frac{1}{2} $, and therefore $\sum_ {p \in P_{large}} \frac{1}{p} \le \frac{1}{2} $

Now consider any $n\in\mathbb{N}$, let Z be the set of integers whose prime factors are all in $P_{small}$

then I can't see how I can carry on and finish it off. Many thanks in advance.

Teymour
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Shenrui Jiang
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    I think in your definition of $P_{\mbox{small}}$ you need to reverse the inequality. It looks like you're trying to reproduce Erdös' proof: https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes – B. Goddard Oct 22 '22 at 16:34
  • $P_{\text{small}} = P_{\text{large}}$? – jjagmath Oct 22 '22 at 17:13

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