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I trying to figure out how to show that:

$a_n = \sin(n)$ (n is Natural number)

Does not converge using Cauchty criterion.

Do you guys have an idea? maybe a Hint?

Thank you.

Bernard
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Rexor
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    If you find many $m,n$ such that $|n/m-2\pi|<C/m^2$, then $|n-2\pi m|<C/m$ is small and therefore $\sin(n)$ close to $\sin(2\pi m)=0$. Doing the same with $\pi/2$, instead of $2\pi$ gives you many $m,n$, with $m$ odd, such that $|n-\pi m/2|<C/m$ and therefore $\sin(n)$ is close to $\sin(\pi m/2)=\pm1$. –  May 11 '18 at 11:16

1 Answers1

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Edit

It appear to be that I miss the fact that OP asked to use Cauchy criterion, this answer prove it in a different way from what OP asked for, I decide to keep it up because it still may help someone.

Supposed the negative: $\lim \sin n=k\in[-1,1]$

Now from that we can conclude that $\lim\cos n=\sqrt{1-k^2}$(why?)

Using that fact we get $2k\sqrt{1-k^2}=\lim\sin 2n=\lim\sin n=k\implies k=\{\pm\frac{\sqrt3}2,0\}$

And $\sqrt{1-k^2}=\lim\cos n=\lim\cos^2 n-\sin^2 n=1-2k^2\implies k=0$

Now we only left to calculate $1-2k^2=\lim\cos(n+1):\quad \lim\cos(n+1)=\lim\cos n\cos 1-\sin n\sin 1=(1-2k^2)\cos 1-k\sin 1$

This gives contradiction to $k=0$

ℋolo
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  • Thank you, but did you use Cauchty criterion? – Rexor May 11 '18 at 11:34
  • @Rexor sorry I didn't saw this. I'll add in the post that it doesn't use it and think on another way in the meantime. – ℋolo May 11 '18 at 11:35
  • @Rexor well, you can take it off from the proof of $k=0$: if $k=0$ then $\lim \cos n=1$ and then using Cauchy: take $\varepsilon=\sin(1/2)$ to get that there exist $N$ such that for all $m,n>N$ we have $|\sin n-\sin m|<\sin(1/2)$, choose $m=n+1$ to get $|\sin n-\sin(n+1)|=2\sin(1/2)|cos(n-1/2)|<\sin(1/2)$ from $\lim \cos n=1$ we have $M$ such that for all $n>M$ we have $1-\varepsilon_1<\cos n$, choose $\varepsilon_1<1/4$ and take $N'>\max{N,M}$ to get that for all $n>N'$ we have $|\sin n-\sin(n+1)|=2\sin(1/2)|\cos(n-1/2)|<1.5\sin(1/2)<\sin(1/2)$, contradiction – ℋolo May 11 '18 at 13:24
  • I wrote this a little messy because it is a comment, if something is not clear ask – ℋolo May 11 '18 at 13:33
  • Why "2sin(1/2)|cos(n−1/2)|<sin(1/2)"? – Rexor May 11 '18 at 14:14
  • @Rexor what is $\sin a-\sin b$? Now use that result with $b=a+1$ and put absolute value. So you get $|\sin n-\sin(n+1)|\color{blue}{=}2\sin(1/2)|\cos(n-1/2)|$, and if $\sin n$ converges it is(the previous expression) less than any $\varepsilon>0$ after some $N$, now choose $\varepsilon=\sin(1/2)$ – ℋolo May 11 '18 at 14:19