More specifically: Is there a function $f$, such that $\sum_{n=1}^\infty a_n$ converges if and only if $|a_n| = O(f(n))$? Or with another Landau symbol like $o(f(n))$?
Since $\sum_{n=1}^\infty \frac 1 {n^\alpha}$ with $\alpha \in \mathbb R$ converges if and only if $\alpha > 1$, we know that $|a_n| = o(1/n)$ is necessary and $|a_n| = O(1/n^{1+\varepsilon})$ for some $\varepsilon > 0$ is sufficient for the convergence of the series $\sum_{n=1}^\infty a_n$.
But of course there are cases, where none of these two conditions apply, like $a_n = \frac{1}{n \log n}$.
Since $\int \frac{1}{x \log x} dx = \log(\log(x)) + C$, we have $\sum_{k=2}^\infty \frac{1}{n \log n} = \infty$ and get the necessary condition $|a_n| = o(1/(n \log n))$.
The integral $\int 1/({x (\log x)^{1+\varepsilon}}) dx = -1/(\varepsilon \log(x)^\varepsilon)$ leads to the sufficient condition $|a_n| = O(1/(n (\log n)^{1+\varepsilon}))$.
We could go on with $\int \frac{1}{x \log(x) \log(\log(x))} dx = \log(\log(\log(x))) + C$ and $\int \frac{1}{x \log(x) \log(\log(x)) \log(\log(\log(x)))} dx = \log(\log(\log(\log(x)))) + C$ etc. to get necessary conditions like $|a_n| = o(1/(\log(x) \log(\log(x)) \log(\log(\log(x)))))$.
So maybe in the limit $$ f(x) = \frac 1 x \prod_{k=1}^\infty \frac{1}{\log^k(x)} $$ could be a candidate such that $|a_n| = o(f(n))$ is necessary and sufficient for the convergence of $\sum_{n=1}^\infty a_n$, but is this true? Does the product converge? Or can this $\log$-$\log$-$\log$-thing never lead to the answer since there are also functions like the iterated logarithm $\log^*(x)$ that grow slower than any iteration $\log^k(x)$? Could we maybe proof, that there is no single function $f$ for a big O notation charaterization of series convergence?
