# Convergence or divergence of the series $\sum\limits_{n = 1}^{\infty} \sin(\pi/n)$

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.

• option $b)$ is true. – DeepSea Aug 29 '15 at 7:19
• @Ganymede : is $d$) wrong ? – user118494 Aug 29 '15 at 7:21
• No....(d) is correct...(b) is also correct... – Empty Aug 29 '15 at 7:22

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, $$\lim_{n\to\infty}\sin\left(\frac\pi n\right)\le\lim_{n\to\infty}\frac\pi n=0$$

$sin(\frac{\pi}{n})$ is asymptotically equivalent to $\frac{\pi}{n}$ so it behaves like the harmonic series which is divergent.

• "equivalent to"? – 6005 Aug 29 '15 at 7:37
• en.wikipedia.org/wiki/Asymptotic_analysis – curiosity Aug 29 '15 at 7:39
• I though that the word "equivalent" in a context like this one has a canonical meaning. Are you ok with "asymptotically equivalent" ? – curiosity Aug 29 '15 at 7:47
• yes, I like that wording better. I think this was a minor point anyway. – 6005 Aug 29 '15 at 7:51
• As long as we know what we're talking about the rest is just vocab ... anyway I've updated my answer – curiosity Aug 29 '15 at 7:55

Hint :

Take , $v_n=\frac{1}{n}$. Then use comparison test.

Since $\sin \frac{\pi}{n} \sim \frac{\pi}{n}$ as $n \to \infty$, and since $\sum_{n \geq 1}\frac{\pi}{n}$ diverges, by the limit comparison test we see that $\sum_{n\geq 1}\sin \frac{\pi}{n}$ diverges.