Nature of the series $\sum\limits_{n}(g_n/p_n)^\alpha$ with $(p_n)$ primes and $(g_n)$ prime gaps

Solution 1:

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 $$ \sum_{n\colon p_n \le x} g_n^2 \ll x^{23/18+\varepsilon}. $$