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I know of Euler's proof that the sum of the reciprocals of the primes diverges. But what if we multiply the primes by it's following prime gap. In other words, is $$\sum_{n=1}^\infty \frac{1}{p_ng_n} = \infty$$ true or false?

Marijn
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    The average gap is $\log p$ by the Prime Number Theorem, I would *guess* that gaps are smaller than $2 \log p$ often enough to cause divergence. – Will Jagy Sep 10 '15 at 20:14
  • See "Prime Gap" in Wikipedia. I think the right answer is "Nobody knows." – DanielWainfleet Sep 10 '15 at 21:05
  • The right answer is: by using the Cauchy-Schwarz inequality we may get rid of the prime gaps, then by exploiting usual density arguments (summation by parts) we may prove the series is divergent. – Jack D'Aurizio Sep 10 '15 at 22:11
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    Related: http://math.stackexchange.com/q/1042649/ – Did Sep 10 '15 at 22:18

1 Answers1

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TRUE: We may get rid of the prime gaps by using Titu's lemma.

We have:

$$ \frac{1}{p_n g_n}+\ldots+\frac{1}{p_N g_N}\geq \frac{\left(\sum_{k=n}^{N}\frac{1}{\sqrt{p_k}}\right)^2}{p_N-p_n}\tag{1}$$ hence if $N$ is around $n^2$ and $n$ is big enough, by partial summation the RHS of $(1)$ is roughly: $$ 4\cdot\frac{p_N+p_n-2\sqrt{p_n p_N}}{(p_N-p_n)(\log N)^2} \tag{2}$$ so by combining $(2)$ with a condensation argument we easily get that the series $\sum_{n\geq 1}\frac{1}{p_n g_n}$ is divergent.

Jack D'Aurizio
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    I don't see the condensation argument. If $N \approx n^2$, then $(2) \asymp \frac{1}{(\log N)^2}$, but continuing this, we get a sequence $N_k \approx x^{2^k}$ for some $x > 1$, and then $\sum_k (\log N_k)^{-2}$ converges. – Dermot Craddock May 24 '25 at 14:53