Primes between gaps of size $nm - n$ for constant $m$ . $\forall n \in \mathbb{N}\; \exists\ m \in \mathbb{N}: \pi(mn) - \pi(n) > 0$, with $\pi(n)$ as the number or primes $ \le n$. 
My hope would be that I can use the following which I just have proven: The sequence $ \pi(n_{i+1}) - \pi(n_i)$ is unbounded $\forall i \in \mathbb{N}$ with $n_{i+1}/n_i \ge 1+\epsilon$, for a certain $\epsilon > 0$ and $\in \mathbb{R}$ (otherwise $\sum_i 1/p_i$ would converge). Can this work?
 A: Yes, you are correct. The fact that $\sum_{s\in S}\frac{1}s$ diverges is sufficient to say that, for any sequence $n_i\in \mathbb N$ such that $\frac{n_{i+1}}{n_i}>1+\varepsilon$, the sequence
$$c_i=|S\cap[n_i,n_{i+1})|$$
is unbounded, where $|\cdot |$ is the size of a set. This is easy to prove since, were such a sequence bounded by some constant $k$, then we'd have that (assuming that $n_0=1$ without loss of generality):
$$\sum_{s\in S}\frac{1}s=\sum_{i=1}^{\infty}\sum_{s\in S\cap [n_i,n_{i+1})}\frac{1}s\leq \sum_{i=1}^{\infty}\sum_{s\in S\cap[n_i,n_{i+1})}\frac{1}{n_i}\leq \sum_{i=1}^{\infty}\frac{k}{n_i}\leq \sum_{i=1}^{\infty}\frac{k}{(1+\varepsilon)^i}<\infty$$
which contradicts that $\sum_{s\in S}\frac{1}s$ diverges.
It's worth noting that the statement that $c_i$ is positive everywhere (which is closer to Betrand's postulate) has almost nothing to do with that sum diverging. For instance, a set like
$$S=\bigcup_{n=1}^{\infty}[n!,2n!]\cap \mathbb N$$
has that $\sum_{s\in S}\frac{1}s$ diverges, but for any sequence $n_i=k^n$ for fixed $k$, the asymptotic density of the $i$ such that $c_i$ is positive is $0$.
