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Can you help me prove that the $\sum \limits _{p \text{ prime}} \dfrac 1 p$ is divergent?

Michael Hardy
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Qwerty
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  • what are your thoughts? – Kushal Bhuyan Nov 21 '15 at 17:48
  • The fact about the series is in Hardy and Wright, I believe due to Mertens. You can look it up as easily as we. Unlikely there is any truly easy derivation – Will Jagy Nov 21 '15 at 17:50
  • HW theorem 427 in the fifth edition. They say Euler proved series divergence in 1737 – Will Jagy Nov 21 '15 at 17:57
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    If $\sum\limits_{n,\in,A} \dfrac 1 n$ converges, then so does $\sum\limits_{n,\in,B} \dfrac 1 n$, where $B$ is the closure of $A$ under multiplication. The closure under multiplication of the set of all prime numbers is all of $\mathbb N$. So the proof of the first assertion I made in this comment is what needs to be done. ${}\qquad{}$ – Michael Hardy Nov 21 '15 at 18:52

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Here is a sketch of a proof. Of course, the hard part lies in the results that are used for this sketch.

A consequence of the Prime Number theorem is the $n$th prime number function $p_n$ is asymptotically equivalent to $\;n\log n$ (actually, it is equivalent to the Prime Number theorem).

Hence $\dfrac1{p_n}\sim \dfrac1{n\log n}$, so that the series $\displaystyle\sum_n\frac1{p_n}\enspace\text{and}\enspace\sum_n\frac1{n\log n}$ both converge or both diverge. However, the latter is a Bertrand's series which diverge.

Bernard
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