Clarification on step in showing sequence $x_{k} = \sin(\frac{k \pi}{3})$ diverges.

Question

There is a specific step in the reasoning of a solution to show that the sequence $x_{k} = \sin(\frac{k \pi}{3})$ diverges that I'm having trouble reconciling. The method used to show this diverges was showing that the sequence is not Cauchy. To do this we took an $\epsilon = 1$ and set $n = 3(2N +1)$ and $m = 6N$. This then implied that $\frac{n \pi}{3} = (2N + 1)\pi$ is an odd multiple of $\pi$ and that $\frac{n \pi}{3} = 2N \pi$ is an even multiple of $\pi$, we conclude $x_{n} = \sin(\frac{n\pi}{3}) = -1$ and $x_{m} = \sin(\frac{m\pi}{3}) = 1$.

It is this last line that has me perplexed. Playing with the multiples of $\pi$ and putting them into the $\sin$ function will always result in $0$. As well none of the multiples of $\frac{\pi}{3}$ will result in a value of $1$ getting obtained. So where are these values coming from?

Answer 1

Yes, the explanation is in error.

Instead, note that

• $x_k={\large{\frac{\sqrt{3}}{2}}}\;$if $k \equiv 1\;(\text{mod}\;6)$.$\\[4pt]$
• $x_k=-{\large{\frac{\sqrt{3}}{2}}}\;$if $k \equiv 5\;(\text{mod}\;6)$.

Then to verify the failure of the Cauchy condition, let $\epsilon=1$, and note that $$|x_{6u+1}-x_{6u+5}|=\sqrt{3} > \epsilon$$ for all integers $u$.