For which $j$ and $r$ does $\sum i^{-2r} \lvert \log (2i) \rvert^{-j}$ converge? I am trying to figure out for which values $j$ and $r$ does
    the series
    $$\sum_{i=2}^{\infty} \frac{1}{i^{2r}|\log(2i)|^{j}}$$
    converges. I have a feeling that given the exponents available,
    I may want to play around with root test to begin to figure out
    these particular values of $j$ and $r$ that will make this converge.
    However, I am still having some issues seeing how this plays into
    the standard theorems on infinite series. I could use some
    assistance on this problem.
 A: First, if $r > -\frac{1}{2}$ then the logarithm grows slowly enough that no power $j$ is large enough to make the series diverge.
Similarly, if $r < -\frac{1}{2}$ then even large negative values of $j$ will not make the logarithm (now in the denominator) grow rapidly enough to alter the fact that the series will diverge.
We are left with the case $r=-\frac{1}{2}$, and in that case, clearly the series diverges (compar to the straight harmonic series) if $j \geq 0$.  That leaves the case of 
$$ \sum_{i=2}^{\infty} \frac{1}{i |\log(2i)|^j} $$ which can be shown to diverge by the same technique that you use to show the harmonic series diverges: By grouping $1, 2, 4, 8 \ldots$ terms you show that the sum, if finite, is larger than the sum of reciprocals of fixed powers of logarithms, which clearly diverges.
A: You should consider using an integral test of convergence, leading you to investigate the integral
$$ \int_2^\infty \frac{1}{x^{2r} \log^j (2x)} dx.$$
This integral converges if $r > \frac{1}{2}$ or if $r = \frac{1}{2}$ and $j > 1$. Otherwise, it diverges.
