Let $ u_{n} = \sin \! \left( \dfrac{\pi}{n} \right) $, where $ n \in \Bbb{N} $, and consider the series $ \displaystyle \sum_{n = 1}^{\infty} u_{n} $. Which of the following is/are true?
(a) $ \displaystyle \sum_{n = 1}^{\infty} u_{n} $ is convergent.
(b) $ \displaystyle \sum_{n = 1}^{\infty} u_{n} $ is divergent.
(c) $ \displaystyle \sum_{n = 1}^{\infty} u_{n} $ is absolutely convergent.
(d) $ u_{n} \to 0 $ as $ n \to \infty $.
Now, $ n \to \infty $ implies $ \dfrac{\pi}{n} \to 0 $, so $ u_{n} = \sin \! \left( \dfrac{\pi}{n} \right) \to 0 $. Also, from the graph of $ \sin $, it looks like this sequence will tend to $ 0 $.
I am not sure about the series options — whether they are all wrong or some are right, and why so.
Best Answer
For $x\le\frac\pi2$, concavity implies $\frac2\pi x\le\sin(x)\le x$.
Therefore, $$ \begin{align} \sum_{n=1}^\infty\sin\left(\frac\pi n\right) &=\sum_{n=2}^\infty\sin\left(\frac\pi n\right)\\ &\ge\frac2\pi\sum_{n=2}^\infty\frac\pi n\\ &=2\sum_{n=2}^\infty\frac1n \end{align} $$ which diverges.
Furthermore, $$ \lim_{n\to\infty}\sin\left(\frac\pi n\right)\le\lim_{n\to\infty}\frac\pi n=0 $$