Stuck at proving whether the sequence is convergent or not

Question

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.

Answer 1

It happens 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.

Answer 2

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.

Answer 3

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 : $X_{n+6}=X_n$

Answer 4

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

Therefore, this sequence diverges.

Answer 5

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?