The sum of the reciprocals of all positive integers is bounded below by $\log(n+1)$; that is, $$\displaystyle\sum_{k=1}^{n}k^{-1}<\log(n+1).$$
The sum of the reciprocals of all prime numbers is bounded below by $\log\left(\log(n+1)\right)$; that is, $$\displaystyle\sum_{p\text{ is a prime with }p\leq n}p^{-1}<\log\left(\log(n+1)\right).$$
So is there a series that increases even slower and is bounded below by $\log\left(\log\left(\log(n+1)\right)\right)$? How about the lower bound $\log^{\circ k}(n+1)$ ( that is, $\log x$ composite with itself $k-1$ times)?
Ideally, given a positive integer $k\geq 2$, I was hoping to find a series $\{a_n\}_{n\geq 1}$ such that $$\log^{\circ k}(n+1)<\displaystyle\sum_{k=1}^{n}a_k<\log^{\circ (k-1)}(n+1).$$
Given $\{b_n\}$ a sequence converging to $+\infty$, you can always find a series whose partial sums grow like $b_n$. It is enough to take $a_n=b_{n+1}-b_n$. Then $\sum_{n=1}^\infty a_n$ is a series whose partial sums are $$ \sum_{k=1}^na_k=b_{n+1}-b_1. $$