The gist of the approach I took is that∑1/p = log(e^∑1/p) = log(∏e^1/p) and logx→ ∞ as x→∞.(adsbygoogle = window.adsbygoogle || []).push({});

Proof outline: let ∑1/p = s(x). (...SO I can write this easily on tablet) and note that e^s(x) diverges since e^1/p > 1 for any p and the infinite product where every term exceeds 1 is divergent. Then loge^s(x) diverges as logs as x→∞ would. Thus, since log(e^s(x)= s(x), the sum is found to be divergent

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# Proof question: the sum of the reciprocals of the primes diverges

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