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Let g(n) be the nth prime gap. (g(1) = 1, g(2) = 2, g(3) = 2, g(4) = 4, etc)

Does the sum of the reciprocals of g(n) converge or diverge as we take n to infinity?

I suspect the problem is equivalent to evaluating the sum of 1/log(n) or something of the sort, but I am afraid I do not know enough about the prime number theorem to answer this question.

Isaac Wachsman
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    Well, if the twin prime conjecture is true, there are infinitely many terms equal to $\frac 12$. – lulu Aug 26 '22 at 21:55
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    More precisely, it is known that there is some gap $g<70,000,000$ that occurs infinitely often, though not a lot can be said about $g$. See, e.g., this. – lulu Aug 26 '22 at 22:02
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    Note: there may well be an easier proof that the sum diverges, but this gets the job done. – lulu Aug 26 '22 at 22:11
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    I think this works as a more elementary sketch: let $T_n$ be the number of primes in $(e^n,e^{n+1})=I$. Then the sum of prime gaps for primes in $I$ is at least, by AM-HM, $\frac{(T_n-1)^2}{(e-1)e^n}$. But $T_n\sim (e-1)e^n/n$ by the PNT, so $\sum_n{\frac{(T_n-1)^2}{(e-1)e^n}}$ diverges, QED. – Aphelli Aug 26 '22 at 23:03
  • We know that infinite many prime gaps do not exceed $246$. This immediately implies that the given sum diverges. – Peter Aug 27 '22 at 08:57

2 Answers2

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One "simple" way of proving divergence here is by knowing that the reciprocal sums of primes diverges, and that g(n)<n. (When I say simple I don't mean that this is an easy way of constructing the proof, but that if you are familiar with the lore, it's an easy way to see divergence at a glance).

Oscar Smith
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One doesn't need to use very much about the primes at all—the bound $p_n = o(n^2)$ suffices:

Proposition: Suppose that $\{a_n\}$ is an increasing sequence of positive real numbers such that $\displaystyle\liminf_{n\to\infty} \frac{a_n}{n^2} = 0$. Then $\displaystyle\sum_{n=1}^\infty \frac1{a_n-a_{n-1}}$ diverges to $\infty$.

Proof: Since $\displaystyle\liminf_{n\to\infty} \frac{a_n}{n^2} = 0$, for any $\varepsilon>0$ we can find $N$ such that $\dfrac{a_N}{N^2} < \varepsilon$. Then at least $\dfrac N2$ of the first $N$ gaps must be less than $2\varepsilon N$ (since the sum of the first $N$ gaps equals $a_N-a_0 < \varepsilon N^2$). It follows that $$ \sum_{n=1}^N \frac1{a_n-a_{n-1}} > \frac N2 \frac1{2\varepsilon N} = \frac1\varepsilon. $$ Since $\varepsilon>0$ was arbitrary, the partial sums of $\displaystyle\sum_{n=1}^\infty \frac1{a_n-a_{n-1}}$ are unbounded.

Greg Martin
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