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Let $p_n$ denote the $n$th prime number and $g_n=p_{n+1}-p_n$ the $n$th prime number gap. This is to ask for which values of $\alpha$ the series $S_\alpha$ converges or diverges, where $$S_\alpha=\sum_n\left(\frac{g_n}{p_n}\right)^\alpha.$$

Context:

  • The obvious lower bound $g_n\geqslant1$ shows that $S_\alpha$ diverges for every $\alpha\leqslant1$.
  • It is known (see the WP page) that $g_n\lt (p_n)^\theta$ for every large enough $n$, for every $\theta\gt\frac34$, hence $S_\alpha$ converges for every $\alpha\gt4$.
  • Various unproven results, such as Cramér's conjecture that $g_n=O\left((\log p_n)^2\right)$, would imply that $S_\alpha$ is finite if and only if $\alpha\gt1$.
  • An answer for $\alpha=2$ would solve (and in fact would be equivalent to a solution of) this other question.

Edit: @GregMartin's answer below yields naturally the more general result that the series $$S_{\alpha,\beta}=\sum_n\frac{g_n^\beta}{p_n^\alpha}$$ converges for every $$\alpha\gt\max\{1,\tfrac5{18}\beta+\tfrac{13}{18}\}.$$ For example, two convergent series are $$\sum_n\frac{g_n^2}{p_n^{4/3}},\qquad\sum_n\frac{g_n^4}{p_n^2}.$$ Actually, an asymptotic control $$ \sum_{n\colon p_n \le x} g_n^2 \leqslant x^{1+\gamma}, $$ for some $\gamma$ in $(0,1)$ (Heath-Brown's result used by @GregMartin being the case of every $\gamma\gt\frac5{18}$) would yield the convergence of $S_{\alpha,\beta}$ for every $(\alpha,\beta)$ such that $$\alpha-1\gt\gamma\cdot(\beta-1)_+.$$

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  • $\begingroup$ what happens if you just replace $g_n$ by $\log p_n?$ $\endgroup$ – Will Jagy Nov 28 '14 at 19:43
  • $\begingroup$ right, simple estimates in Rosser and Schoenfeld (1962) for the substitute series give convergence for your $\alpha > 1.$ So, compare partial sums for the two series up to $p_n,$ could be possible to resolve this. projecteuclid.org/euclid.ijm/1255631807 not suggesting it's easy $\endgroup$ – Will Jagy Nov 28 '14 at 19:53
  • $\begingroup$ @WillJagy "what happens if you just replace g_n by log(p_n)" Then the modified series converges for every α>1, but how is this related to my question? // "simple estimates in Rosser and Schoenfeld (1962) for the substitute series give convergence for your α>1. " Really? Which ones? How? Please be much more specific, I am afraid you lost me. $\endgroup$ – Did Nov 28 '14 at 21:31
  • $\begingroup$ I just meant (3.12) and (3.13), for $n \geq 6$ we get $n \log n < p_n < n (\log n + \log \log n).$ As far as relationship, the logarithm is the expected size of the gap, we can not expect to compare series uniformly, but perhaps something based on averages is worth exploring. I don't have much more, this was never my area. If I get something that works i will leave an answer. $\endgroup$ – Will Jagy Nov 28 '14 at 23:11
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    $\begingroup$ @XuefengMeng Please avoid unnecessary edits to questions of others. The one you suggested, which was, rather mysteriously, approved by two users, was superfetatory since the case $\alpha=1$ is already addressed in the question. $\endgroup$ – Did Jan 9 '15 at 10:40
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The series does converge for every $\alpha>1$.

While our knowledge of individual prime gaps is still somewhat lacking, our knowledge of the gaps on average is rather better. Suppose we have any estimate of the form $$ \sum_{n\colon p_n \le x} g_n^2 \ll x^{2-\delta} \tag{$\ast$} $$ for some $\delta>0$. Then for any $\alpha>1$ we can argue, using Hölder's inequality: \begin{align*} \sum_{n\colon x/2<p_n\le x} \frac{g_n^\alpha}{p_n^\alpha} &\ll \frac1{x^\alpha} \sum_{n\colon x/2<p_n\le x} g_n^\alpha \\ &\le \frac1{x^\alpha} \bigg( \sum_{n\colon x/2<p_n\le x} g_n^2 \bigg)^{\alpha-1} \bigg( \sum_{n\colon x/2<p_n\le x} g_n \bigg)^{2-\alpha} \\ &\ll \frac1{x^\alpha} \big( x^{2-\delta} \big)^{\alpha-1} (x)^{2-\alpha} = x^{-\delta(\alpha-1)}. \end{align*} (We use the fact that the sum of the gaps themselves, of all primes between $a$ and $b$, is just $b-a$ up to one prime gap on each end.) Therefore $$ \sum_{n=1}^\infty \frac{g_n^\alpha}{p_n^\alpha} = \sum_{k=1}^\infty \sum_{n\colon 2^{k-1}<p_n\le 2^k} \frac{g_n^\alpha}{p_n^\alpha} \ll \sum_{k=1}^\infty 2^{-k\delta(\alpha-1)} \ll 1, $$ and so the series converges.

Fortunately, we do know $(\ast)$; indeed, Heath-Brown has established the rather strong version $$ \sum_{n\colon p_n \le x} g_n^2 \ll x^{23/18+\varepsilon}. $$

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  • $\begingroup$ +1. Very nice, thanks a lot. (I guess that, like many number theorists, you use $x_n\ll y_n$ in the Vinogradov sense, that is, to mean that, in Landau's notation, $x_n\in O(y_n)$.) $\endgroup$ – Did Nov 29 '14 at 12:55
  • $\begingroup$ that's right, about $\ll$ $\endgroup$ – Greg Martin Nov 29 '14 at 20:40
  • $\begingroup$ Thanks to (someone) for the bounty! $\endgroup$ – Greg Martin Dec 2 '14 at 21:14
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    $\begingroup$ Here's an alternative argument in proving $(*)$. Let $\theta$ be the exponent in the estimate $g_n \ll p_n^{\theta}$. Then $$\sum_{p_n\leq x}g_n^2 \ll \sum_{p_n\leq x} p_n^{\theta} g_n \ll x^{\theta} \sum_{p_n\leq x} g_n \ll x^{\theta+1} .$$ Any $\theta$ with $0<\theta<1$ will suffice for the purpose of proving $(*)$. $\endgroup$ – Sungjin Kim Jun 17 '17 at 19:10

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