Suppose $\ f:\mathbb{N} \to\mathbb{N}\ $ is increasing and $\ \displaystyle\sum_{n\in\mathbb{N} } \frac{1}{f(n)}\ $ diverges. Then $\ \displaystyle\sum_{n\in\mathbb{N} } \frac{\log (n)}{f(f(n))}\ $ does not necessarily diverge: for example, $\ f(n) = \left\lceil{(n+4)\log(n+4) \log(\log(n+4))}\right\rceil\ $ is a counter-example.
However, I think the following might be true, and if so, I was wondering how to approach proving it:
Suppose $\ f:\mathbb{N} \to\mathbb{N}\ $ is increasing and $\ \displaystyle\sum_{n\in\mathbb{N} } \frac{1}{f(n)}\ $ diverges. Then $\ \displaystyle\sum_{n\in\mathbb{N} } \frac{n^{\alpha}}{f(f(n))}\ $ diverges for each $\ \alpha>0.$
Edit: actually, I can't even think of an $\ f(n)\ $ so that $\ \displaystyle\sum_{n\in\mathbb{N} } \frac{\log^2(n)}{f(f(n))}\ $ converges.
Let's make use of functions with rather uneven growth. Say, $$f(n)=\begin{cases} 2\quad\text{ for }n=1,\\ 4\quad\text{ for }n=2,3,\\ 16\quad\text{ for }n=4\dots15,\\ 256\quad\text{ for }n=16\dots255,\\ \vdots \\ 2^{2^k}\;(2^{2^{k-1}}\leqslant n<2^{2^k})\\ \end{cases}$$ Note that it sends 4 to 16, and 16 to 256, and 256 to its square, and each term dwarfs the previous.
So we have almost $2^{2^k}$ terms of size $2^{2^k}$, hence $\sum\frac{1}{f(n)}$ over these terms is almost 1, so the same sum to infinity diverges. On the other hand, $f(f(n))$ for the same terms is a good deal bigger - I believe, enough so to give you some leeway for your $n^\alpha$.