When do $\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? 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.
 A: 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 $a$.
