The problem:
Show that $$\sum_{n=2}^{\infty} \frac{\sin(\frac{n\pi}{12})}{\ln(n)}$$ converges Conditionally or Absolutely.
Showing that it converges without the absolute value, I have done by using Dirichlet's Theorem. Showing that $\frac{1}{\ln(n)}$ is decreasing and that the partial sum of $\sin(\frac{n\pi}{12})$ is bounded. However how can one show that it converges or diverges in absolute value?
Thank you very much.
Note that infinite sum of (positive) reciprocals of the sequence $(\ln(an+b))_{n\in\mathbb{N}}$ is divergent for all $a, b>0$ (by comparison test or integral test). Since $\sin(\pi+\theta)=-\sin(\theta)$ for any $\theta,$ lets check $\sin\left(\dfrac{n\pi}{12}\right)$ for $n=0, 1, 2, \cdots, 11.$
\begin{array}{|c|c|} \hline n & \sin\left(\dfrac{n\pi}{12}\right) \\ \hline 0, 12 & 0 \\ \hline 1, 11 & \dfrac{\sqrt6-\sqrt2}{4} \\ \hline 2, 10 & \dfrac12 \\ \hline 3, 9 & \dfrac{\sqrt2}2 \\ \hline 4, 8 & \dfrac{\sqrt3}2 \\ \hline 5, 7 & \dfrac{\sqrt6+\sqrt2}{4} \\ \hline 6, 30 & 1 \\ \hline \end{array}
Assuming the absolute convergence we can rearrange the series into $12$ divergent series, according to values in the above table, which is a contradiction.