Show that $\sum_{n\ge{2}}\frac{d(n)}{n\log^2n}$ is divergent.
I've tried to do this using the comparison test, i.e looking for a divergent series with a summand smaller than the summand in the given sum, but the obvious choices ($\frac{1}{n\log^2n}$) seem to converge. Can the Abel test and the divergence of $\sum_{n\ge{2}}d(n)$ (which is determined by the limit as $x\rightarrow\infty$ of the asymptotic expansion) bee used to somehow state that the given sum is divergent? If not what test should be used?
This is a partial answer.
If you prove that you are allowed to use Cauchy's condensation test, you can observe that
$$\sum_{n\geq 2} \frac{2^nd(2^n)}{2^n \log^2 2^n} = \frac{1}{\log^2 2}\sum_{n\geq 2} \frac{n+1}{n^2}$$
is a divergence series, hence your series is divergent.