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I have been trying to determine whether the following sequence is convergent or not. This is what I got:

Exercise 1: Find the $\min,\max,\sup,\inf, \liminf,\limsup$ and determine whether the sequence is convergent or not:

$X_n=\sin\frac{n\pi}{3}-4\cos\frac{n\pi}{3}$

I wrote down a few cases:

$X_1 = \sin\frac{\pi}{3}-4\cos\frac{\pi}{3}= \frac{\sqrt3-4}{2} $
$X_2 = \sin\frac{2\pi}{3}-4\cos\frac{2\pi}{3}= \frac{\sqrt3+4}{2} $
$X_3 = \sin\frac{3\pi}{3}-4\cos\frac{3\pi}{3}=4 $
$X_4 = \sin\frac{4\pi}{3}-4\cos\frac{4\pi}{3}= \frac{4-\sqrt3}{2} $
$X_5 = \sin\frac{5\pi}{3}-4\cos\frac{5\pi}{3}= \frac{4-\sqrt3}{2} $
$X_6 = \sin\frac{6\pi}{3}-4\cos\frac{6\pi}{3}= -4 $

So as found above, $\min = -4$, $\max = 4$, $\inf = -4$, $\sup = 4$, $\liminf = -4$, $\limsup = 4$.

Lets check whether its convergent or not: The sequence is bounded as stated above so lets check if its decreasing or increasing.

$X_n \geq X_{n+1}$

$\sin\frac{n\pi}{3}-4\cos\frac{n\pi}{3} \geq\sin\frac{(n+1)\pi}{3}-4\cos\frac{(n+1)\pi}{3}$

$\sin\frac{n\pi}{3} - \sin\frac{(n+1)\pi}{3} \geq -4\cos\frac{(n+1)\pi}{3} + 4\cos\frac{n\pi}{3}$

I used trigonometrical identity for $\sin\alpha+\sin\beta$ and $\cos\alpha-\cos\beta$ :

$-\cos\frac{\pi(2n+1)}{6} \geq 4\sin\frac{\pi(2n+1)}{6}$

What should I do next? I am stuck here. Thanks, and sorry if I made mistakes.

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  • $\begingroup$ which book is this taken from ? $\endgroup$ – J. Deff Apr 22 at 10:33
  • $\begingroup$ @J.Deff Not sure, why? $\endgroup$ – Exzone Jun 13 at 11:35
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It heppens that your sequence is a cyclic sequence. More precisely, its first six terms are$$-2+\frac{\sqrt3}2,2+\frac{\sqrt3}2,4,2-\frac{\sqrt3}2,-2-\frac{\sqrt3}2,\text{ and }-4$$and then it repeats itself again and again. Therefore:

  • $\max\{X_n\,|\,n\in\mathbb N\}=\sup\{X_n\,|\,n\in\mathbb N\}=\limsup_nX_n=4$;
  • $\min\{X_n\,|\,n\in\mathbb N\}=\inf\{X_n\,|\,n\in\mathbb N\}=\liminf_nX_n=-4$;
  • the sequence diverges.
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  • $\begingroup$ So it is even stupid to check if it converges? It is similar to alternative sequences? $\endgroup$ – Exzone Feb 15 at 15:31
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    $\begingroup$ No, it is not stupid and, yes, it is similar to alternate sequences. $\endgroup$ – José Carlos Santos Feb 15 at 15:32
  • $\begingroup$ Thank you so much. I appreciate it. $\endgroup$ – Exzone Feb 15 at 15:32
  • $\begingroup$ @Exzone from which book is this problem taken from ? $\endgroup$ – J. Deff Apr 22 at 9:36
  • $\begingroup$ @J.Deff You should post this as a comment to the question, not to my answer. $\endgroup$ – José Carlos Santos Apr 22 at 9:49
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The limit of a sequence, if it exists, is equal to its lim inf and lim sup. Accordingly, if the lim inf and lim sup of a sequence are different, then its limit cannot exist.

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Computing the first $6$ terms alone does not prove that $\min X_n=-4$ nor $\max X_n=4$ but rather that $\min X_n\le-4$ and $\max X_n \ge 4$ as it does not say anything about the rest of the sequence.

Hint : $X_{n+6}=X_n$

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Hint: Note that this is a periodic sequence, meaning: $x_{n+6}=x_n$.

Therefore, this sequence diverges.

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    $\begingroup$ periodic alone is not enough to prove that the sequence diverges as constant sequences are periodic and convergent $\endgroup$ – stity Feb 15 at 15:32
  • $\begingroup$ Of course you are right, I wrore only the general idea :) $\endgroup$ – Yarin Luhmany Feb 15 at 15:34
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    $\begingroup$ is there a difference between "diverge" and "does not converge"? $\endgroup$ – costrom Feb 15 at 18:57
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    $\begingroup$ @costrom Typically, diverge is the opposite of converge. It doesn't haven't to fail to converge in any particular way to diverge. $\endgroup$ – Matt Samuel Feb 15 at 23:24
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    $\begingroup$ @Matt I figured as much. it's just a semantic thing, clearly if the sequence tends towards an unbounded value, it "diverges". If I were asked if this particular sequence diverged though, I'm not sure how I'd answer :) $\endgroup$ – costrom Feb 15 at 23:26
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If the sequence is convergent, then all of its subsequences should converge to the same limit. Consider the subsequence defined by $n_{k}=3k$: $$ X_{n_{k}}=\sin(k\pi)-4\cos(k\pi)=-4\cos(k\pi)=\begin{cases} -4 & \text{if }k\text{ is even}\\ +4 & \text{if }k\text{ is odd}. \end{cases} $$ This subsequences oscillates between the values of $-4$ and $+4$. What does that tell you?

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  • $\begingroup$ That subsequences do not converge to the same limit. Thanks! $\endgroup$ – Exzone Feb 15 at 15:35
  • $\begingroup$ Exzone: It tells you that that subsequences does not converge and therefore the original sequence does not converge either. $\endgroup$ – parsiad Feb 15 at 15:36
  • $\begingroup$ Yeah, gotcha! I appreciate your comment. $\endgroup$ – Exzone Feb 15 at 15:37
  • $\begingroup$ *that the subsequence $\endgroup$ – parsiad Feb 15 at 15:43

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