I am working on the exercise of determining when $\sum_{n=1}^{\infty}\frac{\sin n}{n^p}$ converges conditionally. I know if $p\leq0$ it cannot converge by the vanishing condition, and if $p>1$ the series converges absolutely and hence conditionally as well. For the area $0<p<1$, I am completely lost however. I have seen some proofs online that $\sum_{n=1}^{\infty}\frac{\sin n}{n}$ converges conditionally, but this is done using techniques that are way out of my reach (I’m in an introductory analysis class).
The first part of the exercise was to prove that $\sum_{n=1}^{N}\sin n=\frac{\sin(\frac{N+1}{2})\sin(\frac{N}{2})}{\sin(\frac{1}{2})}$, so I think this could be used in some way.
Your are completely right assuming that
$$\sum_{n=1}^{N}\sin n=\frac{\sin(\frac{N+1}{2})\sin(\frac{N}{2})}{\sin(\frac{1}{2})}$$
is an important fact in considering the convergence of the series. In fact this demonstrates that the absolute value of the partial sums of $$ \sum_{n=1}^{N}\sin n $$ is bounded by $\frac{1}{\sin\frac12}$.
Thus by Dirichlet's test the series $\sum_{n=1}^{\infty}\frac{\sin n}{n^p}$ with $p>0$ is convergent.