I've stumbled upon the following result. Let $p_n$ be the $n$-th prime and $\epsilon>0$. Then,
$$ \sum_{n=1}^{\infty}\left(\frac{p_{n+1}-p_n}{p_n}\right)^{1+\epsilon}=C_{\epsilon}<\infty $$
It follows from the above that
$$ \sum_{n=1}^{m}\frac{p_{n+1}-p_n}{p_n}=\ln(p_{m+1})+A+o(1) $$
Does someone have any references on these types of sums concerning the growth rates of primes? Any helpful suggestions on how I can search more efficiently on my own for references would also be very appreciated.
The first sum follows from the following:
Let $f(n)$ be increasing on the intervals $[p_k,p_{k+1})$ and be such that $f(p_n)\sim f(p_{n+1}-1)$. Then,
$$ \sum_{n=1}^{\infty}\frac{p_{n+1}-p_n}{f(p_n)}\geq \sum_{n=1}^{\infty}\sum_{t=p_n}^{p_{n+1}-1}\frac{1}{f(t)}= \sum_{t=p_1}^{\infty}\frac{1}{f(t)}\geq \sum_{n=1}^{\infty}\frac{p_{n+1}-p_n}{f(p_{n+1}-1)}\sim \sum_{n=1}^{\infty}\frac{p_{n+1}-p_n}{f(p_n)} $$
So these sums converge if and only if the others do as well (we made a full circle). Selecting the right function $f(n)$ gives the result. It utilizes the fact that $p_{n+1}<p_n+p_n^{\theta}$ for some $\theta<1$. The smallest theta selectable that is known is $\theta=.525$ but this result follows for any $\theta<1$.