It is well known that $\sum_{n=1}^{\infty} \dfrac1{n^{1+c}} $ and $\sum_{n=1}^{\infty} \dfrac1{n\ln^{1+c}(n)} $ converge for $c > 0$ and diverge for $c \le 0$.
This got me to wondering what would happen if $c$ varied. The obvious choice is $c = a\sin(bn)$ where $b$ is not a multiple of $\pi$.
So that's my question:
When conditions on $a$ and $b$ make $\sum_{n=1}^{\infty} \dfrac1{n^{1+a\sin(bn)}} $ and $\sum_{n=1}^{\infty} \dfrac1{n\ln^{1+a\sin(bn)}(n)} $ converge or diverge?
I don't have a clue.
It is easy to check that, if $b$ is a rational multiple of $\pi$ then the series diverges. So let us focus the case where $b$ is not a rational multiple of $\pi$. We give a slightly general answer.
With any sequence $(a_n)$ and $s_n = a_1 + \cdots + a_n$, we can apply the summation by parts to obtain
$$ \sum_{n=1}^{N} \frac{a_n}{n} = \frac{s_N}{N} + \sum_{n=1}^{N-1} \frac{s_n}{n(n+1)}. $$
Now let $f$ be a $2\pi$-periodic continuous function and set $a_n = n^{-f(bn)}$. Also we assume that there is an interval $J$ in $[0,2\pi]$ such that $f \leq 0$ on $J$. Then we have
$$ \frac{s_n}{n} \geq \frac{1}{n}\sum_{k=1}^{n} \mathbf{1}_J(bn \text{ mod } 2\pi). $$
So if $b$ is not a rational multiple of $\pi$, then by the equidistribution theorem we obtain
$$ \liminf_{n\to\infty} \frac{s_n}{n} \geq \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n} \mathbf{1}_J(bn \text{ mod } 2\pi) = \frac{|J|}{2\pi}. $$
Plugging this back to our summation by parts result,
$$ \sum_{n=1}^{N} \frac{a_n}{n} \geq \sum_{n=1}^{N-1} \frac{\frac{|J|}{2\pi} + o(1)}{n+1} \xrightarrow[N\to\infty]{} \infty. $$
For our case $f(x) = a\sin (x)$, we can pick either $J = [0, \pi]$ or $J = [\pi, 2\pi]$ depending on the sign of $a$.