Nature of the series $\sum\limits_{n}(g_n/p_n)^\alpha$ with $(p_n)$ primes and $(g_n)$ prime gaps 
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 $$\alpha-1\gt\gamma\cdot(\beta-1)_+.$$
 A: 
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
$$
\sum_{n\colon p_n \le x} g_n^2 \ll x^{23/18+\varepsilon}.
$$
