# 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, $$\min = -4$$, $$\max = 4$$, $$\inf = -4$$, $$\sup = 4$$, $$\liminf = -4$$, $$\limsup = 4$$.

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.

• which book is this taken from ? – J. Deff Apr 22 at 10:33
• @J.Deff Not sure, why? – Exzone Jun 13 at 11:35

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.
• So it is even stupid to check if it converges? It is similar to alternative sequences? – Exzone Feb 15 at 15:31
• No, it is not stupid and, yes, it is similar to alternate sequences. – José Carlos Santos Feb 15 at 15:32
• Thank you so much. I appreciate it. – Exzone Feb 15 at 15:32
• @Exzone from which book is this problem taken from ? – J. Deff Apr 22 at 9:36
• @J.Deff You should post this as a comment to the question, not to my answer. – José Carlos Santos Apr 22 at 9:49

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.

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

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

Therefore, this sequence diverges.

• periodic alone is not enough to prove that the sequence diverges as constant sequences are periodic and convergent – stity Feb 15 at 15:32
• Of course you are right, I wrore only the general idea :) – Yarin Luhmany Feb 15 at 15:34
• is there a difference between "diverge" and "does not converge"? – costrom Feb 15 at 18:57
• @costrom Typically, diverge is the opposite of converge. It doesn't haven't to fail to converge in any particular way to diverge. – Matt Samuel Feb 15 at 23:24
• @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 :) – costrom Feb 15 at 23:26

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?

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