Prove by Cauchy theorem that $x_n={\sin {\frac{\pi n}{4}}}$ is a divergent sequence.

Question

The task is to prove that $x_n={\sin {\frac{\pi n}{4}}}$ $(n\in\Bbb{N})$ is not a convergent sequence by Cauchy theorem. The thorem states that a sequence is convergent when $$\forall{\epsilon>0,\quad} \exists{N(\epsilon)\in\Bbb{N},\quad} n>N(\epsilon),\quad \forall{p\in\Bbb{N}}$$ $$|x_{(n+p)}-x_n|<\epsilon$$ In the case of being divergent I guess will have to prove that $$\exists{\epsilon>0,\quad} \forall{N(\epsilon)\in\Bbb{N},\quad} n>N(\epsilon),\quad \exists{p\in\Bbb{N}}$$ $$|x_{(n+p)}-x_n|\ge\epsilon$$ So, I composed the (n+p) term, that is $x_{(n+p)}={\sin {\frac{\pi (n+p)}{4}}}$. Then I wrote $$\Bigl|x_{(n+p)}-x_n\Bigl|=\Bigl|{\sin {\frac{\pi (n+p)}{4}}}-{\sin {\frac{\pi n}{4}}}\Bigl|=$$$$=\Bigl|2\sin \Bigl({{\frac{\pi (n+p)}{4}}-{\frac{\pi n}{4}}\Bigl)}\cos\Bigl({{\frac{\pi (n+p)}{4}}+{\frac{\pi n}{4}}\Bigl)}\Bigl|=$$$$=\Bigl|2\sin \Bigl({\frac{\pi p}{4}}\Bigl)\cos\Bigl({{\frac{\pi n}{2}}+{\frac{\pi p}{4}}\Bigl)}\Bigl|$$ And here I got stuck. I am not sure how to pick a value for p from $\Bbb{N}$, so that I get $|x_{(n+p)}-x_n|\ge\epsilon$ ($\epsilon$ should be a constant I guess). Will appreciate your help and I am sorry if I wrote with mistakes (English is not my native).

Answer 1

I tried again and finally got the answer I wanted. I’ll share it in case someone else needs this approach of solution which works as well. Above I got stuck here :$$=\Bigl|2\sin \Bigl({\frac{\pi p}{4}}\Bigl)\cos\Bigl({{\frac{\pi n}{2}}+{\frac{\pi p}{4}}\Bigl)}\Bigl|$$ As it says that $\exists{p\in\Bbb{N}}$, for proving it will be enough to show that the Cauchy theorem is false for some p. I assumed that p=1 and I got: $$=\Bigl|2\sin \Bigl({\frac{\pi}{4}}\Bigl)\cos\Bigl({{\frac{\pi n}{2}}+{\frac{\pi}{4}}\Bigl)}\Bigl|=$$$$=\Bigl|\sqrt{2}\cos\Bigl({{\frac{\pi n}{2}}+{\frac{\pi}{4}}\Bigl)}\Bigl|$$ As $n\in\Bbb{N}$, $\frac{\pi n}{2}$ can be written as $$1.\pi k + \frac{\pi}{2}, k\in\Bbb{N},\ when \ n=2k+1$$ $$2.\pi k, k\in\Bbb{N},\ when \ n=2k$$ If n=2k+1, then, for instance, $$\Bigl|\cos\Bigl( {{\frac{3\pi}{2}}+\frac{\pi}{4}}\Bigl)\Bigl|=\Bigl|\sin {\frac{\pi}{4}}\Bigl|=\Bigl|\frac{\sqrt {2}}{2}\Bigl|$$ If n=2k, then, for example, $$\Bigl|\cos {\Bigl(\pi+\frac{\pi}{4}\Bigl)}\Bigl|=\Bigl|\cos {\frac{\pi}{4}}\Bigl|=\Bigl|\frac{\sqrt {2}}{2}\Bigl|$$ This means that in any case we will get the same result that is $\Bigl|\frac{\sqrt {2}}{2}\Bigl|$ and we can write $$\Bigl|\sqrt{2}\cos {\frac{\pi}{4}}\Bigl|=1$$ This is the same if we write $$\Bigl|x_{(n+p)}-x_n\Bigl|=1$$ which means we showed that $\exists{\epsilon\ge {0}}$ that $$\Bigl|x_{(n+p)}-x_n\Bigl|\ge {\epsilon}=1$$