I must determine whether if the following series converges, converges absolutely, or diverges: $$\sum_{n=1}^\infty\sin(n)\sin\left(\frac{\pi}{2n}\right)$$ By the comparison test, I have already found that $\sum\limits_{n=1}^\infty \left(\sin\left(\frac{\pi}{2n}\right)\right)^p$ converges for $p>1$ and diverges for $p \leq 1$. Thus, $
\sum\limits_{n=1}^\infty\sin\left(\frac{\pi}{2n}\right)$ diverges by this criterion. I suspect the entire series will also diverge, and that I have to use the comparison test, but I encountered an issue:
Since $-1 \leq \sin n \leq 1$, we have that $\sin(n)\sin\left(\frac{\pi}{2n}\right) \leq \sin\left(\frac{\pi}{2n}\right)$. This would be useful if the series represented by the term on the right converged; in its current state, this cannot be used to prove divergence.
Is my reasoning wrong? Should I be using another test for this series? Thank you.
Best Answer
Since the partial sums of $\sum_{n}\sin(n)$ are bounded and $\sin(\frac{\pi}{2n})$ is a decreasing sequence that goes to $0$, Dirichlet test proves the convergence of your series.
To get a bound on the partial sums of $\sum_{n}\sin(n)$, note that $$ \left|\sum_{k=0}^n \sin(k)\right|=\left|\Im\sum_{k=0}^ne^{ik}\right|\leq\left|\frac{1-e^{i(n+1)}}{1-e^i}\right|\leq \frac{2}{|1-e^i|}$$