Question on convergence / divergence of sums of reciprocals of positive integers

Studying the most common known convergence tests, I found that in some cases such tests are inconclusive (e.g., ratio test when $$r=1$$, comparison test when the series is conditionally convergent, etc), so I have been thinking of some possible convergence test for series involving sums of reciprocals of positive integers.

The rational behind this test is the following: somehow, the density of given subsets of positive integers can be evaluated and compared through their partial sums. For instance, it is intuitive that a set of $$n$$ positive integers such that $$\sum_{k=1}^{n}a_{k}=\frac{n(n+1)}{2}$$ is more dense than a set of $$n$$ positive integers such that $$\sum_{k=1}^{n}b_{k}=\frac{n(n+1)(n+2)}{6}$$.

Taking a look at some of the most known series of reciprocals of positive integers, it is easy to see that precisely $$\sum_{k=1}^{n}a_{k}=\frac{n(n+1)}{2}$$ is the partial sum corresponding to the most dense possible subset of positive integers, as it is the sum of consecutive positive integers starting at $$1$$. It is known and easily provable that the sum $$\sum_{k=1}^{\infty}\frac{1}{a_{k}}$$ diverges, at a rate of aproximately $$\ln(n)$$. Other known diverging sequence, the sum of reciprocals of prime numbers, diverges at a rate of aproximately $$\ln\ln(n)$$, and the partial sum of consecutive prime numbers is aproximately $$\sum_{k=1}^{n}p_k=\frac{1}{2}n^2\ln(n)$$. However, the already noted partial sum $$\sum_{k=1}^{n}b_{k}=\frac{n(n+1)(n+2)}{6}$$ corresponds to the set of triangular numbers, and we have that $$\sum_{k=1}^{\infty}\frac{1}{b_{k}}=2$$.

The possible convergence test stated relies on the existence of some function $$F(n)$$, bounded as $$\frac{1}{2}n^2\ln(n), such that for every infinite subset of positive integers $$S=\left\{ a_{1},a_{2},...\right\}$$ such that $$\lim_{n\rightarrow\infty}\frac{\sum_{k=1}^{n}a_{k}}{F(n)}=\infty$$, then we can affirm that $$\sum_{a\in S}\frac{1}{a}<\infty$$; and if $$\lim_{n\rightarrow\infty}\frac{\sum_{k=1}^{n}a_{k}}{F(n)}=0$$, then we can affirm that $$\sum_{a\in S}\frac{1}{a}=\infty$$.

Therefore, the test would be based on the sum of the denominators of the sequence, and would have the following form:

(Possible) Convergence test

Given some infinite subset of positive integers $$S=\left\{ a_{1},a_{2},...\right\}$$ such that $$\lim_{n\rightarrow\infty}\frac{\sum_{k=1}^{n}a_{k}}{F(n)}=\infty$$, then we can affirm that $$\sum_{a\in S}\frac{1}{a}<\infty$$; and if $$\lim_{n\rightarrow\infty}\frac{\sum_{k=1}^{n}a_{k}}{F(n)}=0$$, then we can affirm that $$\sum_{a\in S}\frac{1}{a}=\infty$$

The question now is: is it possible the existence of such function $$F(n)$$? Is it compatible with the fact proved here: https://math.stackexchange.com/questions/452053/is-there-a-slowest-rate-of-divergence-of-a-series#:~:text=Talking%20about%20getting%20closer%20to,%22the%20slowest%20diverging%20series%22?

I believe it is possible the existence of such a function, and that it would be compatible if it did not exist any partial sum of positive integers equal to $$F(n)$$. For example, if hypotetically $$F(n)=n^e$$, there would not exist any set of positive integers such that the rate of convergence/divergence were $$0$$.

Any comment / guess on how to 1) prove the existence or non-existence of $$F(n)$$, and 2) approximating $$F(n)$$ would be welcomed!

Unfortunately, even a fast growing function $$F(n)$$ fails to assure $$1/a_n\to 0$$. For instance, put $$a_{2k}=k!$$ and $$a_{2k+1}=1$$ for each natural $$k$$. Even when we require that $$\{a_n\}$$ is non-decreasing, it fast growth can fail to assure convergence of a series $$\sum_{i=1}^n \tfrac 1{a_i}$$. For instance, for each very fast increasing function $$g:\Bbb N\to\Bbb N$$ let the sequence $$\{a_n\}$$ consists of consecutive blocks of number $$g(k)$$ and length $$g(k)$$. Then a sequence $$\{1/a_n\}$$ diverges, but a sequence $$\{\sum_{i=1}^{n} a_i\}$$ has big jumps at $$g(k+1)$$ at each $$n(k)=1+\sum_{i=1}^k g(i)^2$$.
On the other hand, the inequality between arithmetic and harmonic means implies that $$\sum_{i=1}^n \frac 1{a_i}\ge \frac{n^2}{\sum_{i=1}^n a_i},$$ thus if the right-hand side of this inequality is unbounded then the series $$\sum_{i=1}^n \tfrac 1{a_i}$$ diverges.