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Looking to see if I am doing this proof correctly. Any help would be appreciated.

Consider the sequence {$x_n$}$_{n=1}^\infty$ where

$x_n$ = $\left(\frac{ncos(n)-23}{n}\right)$

This is not a convergent sequence. Prove that it has a convergent subsequence. Do it by making use of the appropriate theorem, and be sure to confirm that the hypotheses of that theorem holds (by making use of the triangle inequality).

My solution:

Proof: $\left(\frac{ncos(n)-23}{n}\right)$ = cos(n) - $\left(\frac{23}{n}\right)$. This is a bounded sequence (I think). Since $x_n$ is a bounded sequence, it follows by the Bolzano-Weierstrass Theorem that $x_n$ has a convergent subsequence. I'm not sure where to start for confirming it using the triangle inequality though.

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  • $\begingroup$ Start with $-1 \leq \cos{n} \leq 1$ $\endgroup$ – rtybase Sep 28 '16 at 19:37
  • $\begingroup$ show it only for cos(n), the other is convergent. $\endgroup$ – hamam_Abdallah Sep 28 '16 at 19:40
  • $\begingroup$ Any bounded sequence has a convergent subsequence, and your sequence is clearly bounded. $\endgroup$ – Jack D'Aurizio Sep 28 '16 at 19:52
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It follows directly from $|\cos(n)|\le 1$, $\frac{a}{n}\le a$ for any $a>0$, $n\in \mathbb N_{>0}$. With these two, it is clear that $$\left |\frac{n \cos (n)-23}{n} \right |=\left |\cos(n)-\frac{23}{n}\right |\le |\cos(n)|+\frac{23}{n}\le 1+23=24$$

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  • $\begingroup$ What would I take big N to be in this case when going by the formal definition of convergence? Are you proving boundedness or convergence? $\endgroup$ – Remy Sep 28 '16 at 19:47
  • $\begingroup$ @JohnH Could you please clarify? Bolzano-Weierstrass theorem tells us that if $(x_n)_{n\in \mathbb N}$ is a bounded sequence in $\mathbb R$, then there exists a convergent subsequence $x_{\varphi(n)}$ of $(x_n)_{n\in \mathbb N}$. We have shown that $(x_n)_{n\in \mathbb N}$ is a bounded sequence, hence the theorem holds and we have proven the existence of such a subsequence. $\endgroup$ – Lonidard Sep 28 '16 at 19:51
  • $\begingroup$ Oh ok so you proved it is bounded, correct? Which allows us, in turn, to deduce that is has a convergent subsequence (by the Bolzano-Weierstrass Theorem)? $\endgroup$ – Remy Sep 28 '16 at 19:54
  • $\begingroup$ @JohnH Correct. $\endgroup$ – Lonidard Sep 28 '16 at 20:50
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Note that $$ x_n=\frac{n\cos(n)-23}{n}=\cos(n)-\frac{23}{n} $$ so $$ -2\le x_n\le 1 $$ for $n\ge23$. Therefore the sequence is bounded.

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  • $\begingroup$ @JohnH For $n\ge23$. :) $\endgroup$ – egreg Sep 28 '16 at 19:50
  • $\begingroup$ Oh ok that makes sense now haha $\endgroup$ – Remy Sep 28 '16 at 19:56
  • $\begingroup$ @JohnH In every such situation, you just need to prove that the sequence is bounded from some point on. Use the point where the proof becomes easier. But, of course, you can give a larger bound, no problem. $\endgroup$ – egreg Sep 28 '16 at 19:58
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Since $x_n = \cos n - 23/n$ and $23/n \to 0,$ $x_n$ has as many subsequential limits as does $\cos n.$ It is well known that $\{\cos n: n\in \mathbb N\}$ is dense in $[-1,1].$ It follows that for every $x\in [-1,1],$ there is a subsequence of $x_n$ that converges to $x.$

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