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 $\frac{1}{2}n^2\ln(n)<F(n)<\frac{n(n+1)(n+2)}{6}$, 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!
Best Answer
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.