How can i prove that the sequence $f_n(x)=cos(nx)$ does not converge for $x \neq 0$? I tried to prove that is not a Cauchy sequence without success.
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Did you try the case when $x=\pi$? – Dec 11 '16 at 16:32
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Can you get something from here: http://math.stackexchange.com/q/24898/9464? Not an exact duplicate though, they are almost the same. – Dec 11 '16 at 16:34
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Yes, for x=π its ok..but how can i prove it for all x $ \neq0$? – Millis Dec 11 '16 at 16:34
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Look at $f_{2n}(x) - f_n(x)$. – Daniel Fischer Dec 11 '16 at 16:41
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$(\cos(nx))^{'}=-n\sin(nx).$Now $-n\sin(nx)=0 $ implies $x=m \pi \hspace{1mm} \text{where} \hspace{1mm} m\in \mathbb{Z}$.Which shows that we have extremas at these points. We will have then have at $x = (2m+1)\pi , $ $f_{2k} = 1$ and $f_{2k+1} = -1 $ hence $f_n(x)$ does not converge for $x \neq 0$.
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Suppose $0<x<\pi.$ Let $A= \{e^{int}: -x/2\le t \le x/2\}.$ Then $e^{inx} \in A$ for infinitely many $n.$* This implies $\cos (nx) \ge \cos (x/2)>0$ for infinitely many $n.$ Similarly, $\cos (nx) \le -\cos (x/2)<0$ for infinitely many $n.$ It follows that $\cos nx$ has a subsequence whose consecutive terms are at least $2\cos(x/2)$ apart. Hence this subsequence diverges, which implies $\cos nx$ diverges. This in turn implies $\cos (nx)$ diverges for all $x\notin \pi\mathbb Z.$
$*$ This is because $e^{inx}, n=1,2,\dots$ marches around the circle infinitely many times in steps of arc length $x.$ The arc $A$ has arc length $x.$ Thus there is no way $e^{inx}$ can "jump over" $A$ in this process, so it will land in $A$ at least once in every orbit of the circle, i.e., at least once every $\lceil 2\pi/x \rceil$ steps.
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