I suppose that yes, it does give some insight into the nature of the primes; or rather, their distribution amongst the natural numbers. Realize that Zeta( *a* ) is finite for all *a* > 1, and this means that given *any* positive real *a* greater than 1 will produce a finite value of $\sum_n^\infty=\frac{1}{n^a}$ (the closer to 1 it gets, the larger it will get). However, the partial sums of the reciprocals of the primes *do not converge* .
What does this mean? Well, consider that there exists some exponent *j* and coefficient *k* such that, for every integer *n* , $\frac{k}{n^j}\geq\frac{1}{p(n)}$, where *p* ( *n* ) is the *n* ^{th} . If we flip the reciprocals, we also flip the direction of the inequality, and arrive at $\frac{n^j}{k}\leq p(n)$, or $n^j\leq p(n)\cdot k$. This means that, given j,k such that the above conditions hold, the function *p* ( *n* ) grows *at least as fast as* this function with exponent *j* .
Looking back at our $\frac{k}{n^j}$, we examine $\sum_{n=1}^{\infty}\frac{k}{n^j}=k\cdot \zeta(n)\lt\infty$, as per the finite nature of *k* and the Zeta function at values greater than 1. But, the *n* ^{th} term in the series is always greater than the *n* ^{th} prime number, and therefore the whole sum is therefore greater than the sum of the reciprocals of the primes, which, as we know, is infinite.
This creates a contradiction, in that the sum, which we know is finite, is greater than an infinite series. Therefore, there can't exist a *j* and *k* that satisfy $\frac{k}{n^j}\geq\frac{1}{p(n)}$ for all *n* , which means that the prime numbers themselves grow slower than any power.

All this does is it says the primes grow slower than any power, but it doesn't say *how* they grow, which the Prime number theorem *does* (and supports this conclusion). I suppose this could be generalized to say that any function which can be shown to, for all sufficiently large terms, grow slower than any power greater than 1, it is divergent (the primes being an example, and the evens being another).