What I claim is that, letting $p_n$ being the nth prime $\frac{\lvert\{p_n\leq N\mid p_{n+1}-p_n\leq \log^{1-\epsilon} N\}\rvert}{N/\log N}\rightarrow 0$

in other words the proportion of the gaps between primes up to N which are smaller than $\log ^{1-\epsilon} N$ is tending to 0. Thus although we know that there are infinitely many bounded gaps, we see here that they are almost all of size at least logarithmic.

To prove it, we recall a classical application of Selberg’s sieve: $\pi_m(x):=\lvert\{p \leq x\mid p+m\text{ is also a prime}\}\rvert \ll \frac{x}{\log^2 x}\prod_{p\mid m}(1+1/p)$

where the implied constant is absolute (does not depend on m).

Now this product is obviously less than than $\prod_{p\leq \omega(N)}(1+1/p)$ which by Mertens’s theorem and the obvious bound $\omega (n)\ll \log n$ implies that is at most $\log\log N$. But now the numerator of our fraction is obviously $\displaystyle{\sum_{m\leq \log^{1-\epsilon} N}\pi_m(N)\ll\frac{N}{\log^2 N} \sum_{m\leq \log^{1-\epsilon} N} \log\log m\ll \frac{N}{\log^2 N}\log^{1-\epsilon/2}N}$

The last term is simply $\frac{N}{\log^{1+\epsilon/2}N}$ which is indeed negligible compared to the number of primes until N, whence the result.

Notice that we could have argued a bit differently: indeed, $\sum_{m\leq M}\prod_{p\mid m}(1+1/p)=\sum_{d}[M/d]\frac{\mu(d)^2}{d}\ll M$ (instead of $M\log\log M$ previously). Thus, the number of primes until N such that the next prime is at distance at most M is bounded by $\frac{N}{(\log N)^2}M$, which is $o(1)$ as soon as $M=o(\log N)$.

As a conclusion, gaps of size $p_{n+1}-p_n=o(\log p_n)$ have density 0.