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.
 A: Extrapolating off of Marvis's work in the comments section, we see that if $b(n)$ is similar to $\dfrac{1}{\log(n)^{k}}$ for a fixed $k\ge 2$, ($k$ exponent here) then the series will converge. Furthermore, if $b(n)$ is similar to $\dfrac{1}{\log(n)}$ we observe divergence. Consider extending this process to a fixed $j$ so that if $b(n)$ is similar to $\dfrac{1}{\log n\log\log n \log\log\log n ...\log^{(j)} n}$, we have divergence. These are mere examples. 
In general, 
\begin{align*}
\sum_{n=2}^{\infty} \log(n)(b(n) - b(n+1)) = \sum_{n=2}^{\infty} \log(n)b(n) - \sum_{n=2}^{\infty} \log(n)b(n+1) 
\end{align*}
If $b(n)$ is a non-decreasing function then 
\begin{align*}
b(n) \le b(n+1) \Rightarrow \\
\mbox{ For $n\ge2$ we have } \log(n)b(n) \le \log(n)b(n+1) \Rightarrow \\
\log(n)b(n)-\log(n)b(n+1) \le 0 \\ 
\end{align*}
Now $b'(n) \ge 0$. 
If $b'(n) = 0$, then the entire sum is $0$. (Uninteresting) 
If $b'(n) > 0$ (Increasing) and $b''(n)>0$ (Second Derivative also increasing), then for $n >>2$ 
$\log(n)b(n)-\log(n)b(n+1)$ ~ $-\log(n)b(n+1)$ 
$b(n) -b(n+1)$ can get arbitrarily large because the vertical distance between $b(n)$ and $b(n+1)$ approaches $\infty$. This is because for any interval, $[n,n+1]$, we have this by the MVT. 
So it would stand within reason that the sum would tend to $-\infty$. 
What about when $b''(n) < 0 $  but $b'(n) > 0 $ $\forall n$ ? 
Well the derivative is decreasing implying that for any interval $[n,n+1]$, the vertical distance between $b(n)$ and $b(n+1)$ is decreasing by MVT, so the sum would eventually converge. 
A: 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.
