For what real $r$ does $\sum_{n=2}^\infty\frac1{n(\ln n)^{1+ir}}$ converge and for which $r$ does the sum diverge? A standard convergence result is that
$\sum_{n=2}^\infty\frac1{n(\ln n)^p}$
converges for real
$p > 1$
and diverges for
$p \le 1$.
How about complex $p$
with real part $1$?
Specifically,
for what real $r$
does
$\sum_{n=2}^\infty\frac1{n(\ln n)^{1+ir}}$
converge and for which $r$ does the sum diverge?
How about the usual series of sums
$\sum_{n=e_2}^\infty\frac1{n(\ln n)(\ln  \ln n)^{1+ir}}$
, ...,
$\sum_{n=e_k}^\infty\frac1{n(\ln n)(\ln \ln n)...(\ln ...  \ln n)^{1+ir}}$,
where $e_k$ is large enough so that
the iterated $\ln$s all are positive?
I have not made any progress on this question
because of my lack of knowledge
of complex analysis.
However, I guess that
the cases when
$r = \pi s$ with
$s$ rational and irrational
might have significantly different answers.
 A: Consider the series $$\sum_{n=2}^\infty \dfrac{1}{n (\ln n)^{1+ir}} = \sum_{n=2}^\infty \dfrac{1}{n \ln n} \exp(-i r \ln \ln n)$$
where $r > 0$ ($r < 0$ is similar).  Note that $\sin(r \ln \ln n) > 1/2$ 
when $(2 k + 1/6) \pi < r \ln \ln n < (2 k + 5/6) \pi$ for some integer $k$.
Let's say this is for $M_k < n < N_k$.  Now
$$ \sum_{M_k < n < N_k} \dfrac{1}{n \ln n} \approx \ln \ln (N_k) - \ln \ln (M_k) \approx \left(2k+ \frac{5}{6}\right) \frac{\pi}{r} - \left(2k + \frac{1}{6}\right)\frac{\pi}{r} = \dfrac{2\pi}{3r}$$ 
Thus there are  $M_k < N_k$ with $M_k$ arbitrarily large at which the imaginary parts of the partial sums of your series differ by at least a positive constant.  We conclude that the series diverges.
A: Consider instead the integral $ \int_2^{\infty} \frac{1}{n \log (n)^{1+ir}} \,dn$. 
It is equal to $\lim_{n \rightarrow \infty} (- \frac{\log(n)^{-ir}}{ir} + \frac{\log(2)^{ir}}{ir})$. Since $\log(n) \rightarrow \infty$ we have that $\log(n)^{-ir} = \exp(-ir \log(\log(n)) = \cos(-r \log(\log(n)) + i\sin(-r \log(\log(n))$ which diverges when $r \ne 0$.$~$So by the integral test the original series diverges.
A similar method can be applied to the more complex sums. The general antiderivative can be evaluated through recursion
: $ \int \frac{1}{n \log(n)\log(\log(n)) \dots \log(\dots(\log(n)))^{1+ir}} \,dn = \frac{\log(\dots(\log(n)))^{-ir}}{-ir}$ In this way the other series can be shown to diverge.
