<s>[INST] Stuck at proving whether the sequence is convergent or not 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, $\min = -4$, $\max = 4$, $\inf = -4$, $\sup = 4$, $\liminf = -4$, $\limsup = 4$.
Let's 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.
 [/INST] 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 subsequence oscillates between the values of $-4$ and $+4$.
What does that tell you?
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