sequences and series - Convergence of $sum_{i=2}^{infty}frac{1}{i cdot f(i)}$

[Edited to fix typo]

Is there a precise formulation for when the sum

$$
\sum_{i=2}^{\infty}\frac{1}{i \cdot f(i)}
$$ converges, in terms of the function $f$? Assume that $f$ is smooth and monotonically increasing.

If $f(i) \gtrsim i^c$ for any $c>0$ then we know it converges. If $f(i)$ is a constant then we know it doesn't. We can try functions in between. For example setting $f(i) = 2^{\sqrt{\log(i)}}$ makes the sum converge but setting $f(i) = \log(i)$ makes it diverge according to Wolfram Alpha

There are of course a lot of functions so it might be hard to write a full classification. How about if we only including elementary functions that, for example, use only powers and logs?

Update. Is something like the following conjecture true? Consider $\sum_{i=\ell}^{\infty}\frac{1}{i \cdot f(i)}$ and set $\ell$ to be the smallest positive integer so that $f(\ell) >0$. The sum converges if and only if there exists $c>0$ such that $f(i) \gtrsim c \log(i)\log{\log(i)}\log{\log{\log(i)}}\dots$ where the $\log$ is applied an (as yet) unknown but fixed number of times.

Answer

Since $f$ is monotonically increasing, we can use the integral test.

Define repeated composition by
$$
f^{\circ0}(x)=x\quad\text{and}\quad f^{\circ k+1}(x)=f\circ f^{\circ k}(x)
$$

Note that if
$$
f_n(x)=\prod_{k=1}^n\log^{\circ k}(x)

$$
then, for $n>0$,
$$
\begin{align}
\int_{\exp^{\circ n}(1)}^\infty\frac{\mathrm{d}x}{xf_n(x)\log^{\circ n}(x)^a}
&\stackrel{\hphantom{\text{induction}}}=\int_{\exp^{\circ n}(1)}^\infty\frac{\mathrm{d}\log(x)}{f_n(x)\log^{\circ n}(x)^a}\\
&\stackrel{\substack{x\mapsto e^x\\\hphantom{\text{induction}}}}=\int_{\exp^{\circ n-1}(1)}^\infty\frac{\mathrm{d}x}{xf_{n-1}(x)\log^{\circ n-1}(x)^a}\\
&\stackrel{\text{induction}}=\int_1^\infty\frac{\mathrm{d}x}{x^{a+1}}\\
&\stackrel{\hphantom{\text{induction}}}=\frac1a
\end{align}

$$
Therefore, for all $n\ge0$,
$$
\int_{\exp^{\circ n}(1)}^\infty\frac{\mathrm{d}x}{xf_n(x)}
$$
diverges, yet for any $a>0$,
$$
\int_{\exp^{\circ n}(1)}^\infty\frac{\mathrm{d}x}{xf_n(x)\log^{\circ n}(x)^a}
$$
converges. As far as logs and powers go, these border convergence/divergence pretty closely.