# Does $\sum\limits_{n=1}^\infty\sin(n)\sin\left(\frac{\pi}{2n}\right)$ converge?

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.

• It certainly can't converge absolutely (and that should be straightforward to prove), but convergence of this series looks remarkably tricky at least at first glance... – Steven Stadnicki Apr 1 '15 at 3:48
• Dirichlet test does the trick. – Gabriel Romon Apr 1 '15 at 3:49

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 $$\left|\sum_{k=0}^n \sin(k)\right|=\left|\Im\sum_{k=0}^ne^{ik}\right|\leq\left|\frac{1-e^{i(n+1)}}{1-e^i}\right|\leq \frac{2}{|1-e^i|}$$
• How would I now prove that the series is not absolutely convergent? For $\sum_{n=1}^{\infty} \lvert \sin n \rvert \sin\left(\frac{\pi}{2n}\right)$, we have that $\lvert \sin n \rvert \sin\left(\frac{\pi}{2n}\right) \leq \sin\left(\frac{\pi}{2n}\right)$, but again, this cannot be used. – Douglas Fir Apr 1 '15 at 4:43
• This is a different question. Use the inequality $\forall x\in [-\pi/2,\pi/2], |\sin(x)|\geq \frac{2}{\pi}|x|$ and the divergence of the series $\sum \frac{|\sin(n)|}{n}$ (see math.stackexchange.com/questions/264980/…) – Gabriel Romon Apr 1 '15 at 4:47