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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.

Marcelo
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1 Answers1

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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 $$ \left|\sum_{k=0}^n \sin(k)\right|=\left|\Im\sum_{k=0}^ne^{ik}\right|\leq\left|\frac{1-e^{i(n+1)}}{1-e^i}\right|\leq \frac{2}{|1-e^i|}$$

Michael Hardy
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Gabriel Romon
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  • 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. –  Apr 01 '15 at 04:43
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    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 http://math.stackexchange.com/questions/264980/how-to-prove-that-sum-n-in-mathbbn-frac-sin-nn-diverges) – Gabriel Romon Apr 01 '15 at 04:47