Prove that the sequence $\sin\left(\frac{n\pi}{3}\right)$ diverges I don't want to hear that since $\sin$ is a periodic function, etc, then we are done. I would like to see a simple proof that make use of the definition of convergence of a sequence.
I have tried to assume (by contradiction) that $\lim \left(\sin\left(\frac{n\pi}{3}\right) \right) = a$. But then I don't know what to do. In the examples I have seen, they assume that $\epsilon$ is equals to some value, but I don
t understand where this value of $\epsilon$ comes from. 
 A: Put 
$$a_n:=\sin\frac{n\pi}3\implies \begin{cases}a_{3n}=0\xrightarrow[n\to\infty]{}0\\{}\\a_{6n+1}=\frac{\sqrt3}2\xrightarrow[n\to\infty]{}\frac{\sqrt3}2\end{cases}$$
so we have two subsequences converging to different limits and thus the original sequence's limit cannot exist.
A: Observing that $a_{6k+1}=\sqrt3/2$ and $a_{6k+4}=-\sqrt3/2$, we can prove directly from the definition that there is no number that qualifies as its limit. (And thus, by definition, it diverges).
Every $L\ge0$ is a non-limit: Set $\varepsilon=\sqrt3/4$ and look for an $N$ such that $|a_n-L|<\varepsilon$ for all $n>N$. That is not possible because you always have $6N+4>N$ and $a_{6N+4}=-\sqrt3/2$ is too far from $L$.
Similarly every $L\le0$ is a non-limit -- with exactly the same argumemt except we take $a_{6N+1}$ instead of $a_{6N+4}$.
(The exact value of $\varepsilon$ used here is not important. It just has to be some number greater than $0$ and smaller than $\sqrt3/2$. The latter condition is because we want $|L-(-\sqrt3/2))|=L+\sqrt3/2$ to be strictly larger than $\varepsilon$ for every $L\ge 0$).
A: Start by assuming that $L$ exists. The take $n>n_0$ such that $\sin \frac{\pi n}{3}=0$ and $|\sin \frac{\pi n}{3}-L|<0.01$. By definition all $f(n > n_0)$ must be within this radius from $L$. Now use periodicity of $\sin$ and show it doesn't hold for $n+1$. Since we took it for arbitrary $n$, $L$ doesn't exist. 
A: If you're not sure where $\epsilon$ is coming from, you need to look back over the definition of a limit.
A sequence $a_n$ is said to converge to a  limit $L$ if, for any $\epsilon > 0$, there exists some natural number $N$ such that:
$$\text{whenever }n > N, \text{ we have } |a_n - L| < \epsilon.$$
If we want to show our sequence doesn't converge to $L$, then all we need is to find is some $\epsilon > 0$ for which there is no natural number $N$ for which $|a_n - L| < \epsilon$ even though $n > N$.
The key idea is that a certain statement is supposed to hold (namely, for all $\epsilon > 0$, there's some $N$, such that...). We need to show that it doesn't hold; that there's at least one $\epsilon > 0$ that doesn't cooperate.
EDIT: In this case, we can make an educated guess about what values of $\epsilon$ are likely to get the job down. Notice that, for your sequence, $a_n \in \left\{0, \dfrac{\sqrt{3}}{2},\dfrac{-\sqrt{3}}{2}\right\}$. Intuitively, your limit should be one of these terms, and the possible absolute differences between these terms are $0, \dfrac{\sqrt{3}}{2}$, and $\sqrt{3}$. If you just pick an $\epsilon$ smaller than the smallest difference here (say, $\epsilon = \dfrac{\sqrt{3}}{4}$, although many more potential $\epsilon$ would work), then you should be able to show $|a_n - L|$ will be greater than this $\epsilon$ infinitely often, no matter what $L$ is.
A: The proof is in the second part of my post. First, let's analyze the problem. Let $(a_n)$ be this sequence. It converges to some $a$, if and only if for all $\varepsilon > 0$, there exists an integer $N \geq 0$ such that, for all $n \geq N$, $|a_n-a| < \varepsilon$.
In particular, if the sequence converges, then for all $n$, $m \geq N$,
$|a_n-a_m| \leq |a_n-a|+|a-a_m| < 2 \varepsilon.$
Conversely, if there exists $\varepsilon > 0$ such that, for all $N \geq 0$, for all $n$, $m \geq N$,
$|a_n-a_m| \geq 2 \varepsilon,$
then the sequence does not converge.
Now, let us look at the sequence $(a_n)$. What does it look like? Well, its values are $0$, $\sqrt{3}/2$, $\sqrt{3}/2$, $0$, $-\sqrt{3}/2$, $-\sqrt{3}/2$, $0$, $\sqrt{3}/2$, $\sqrt{3}/2$, $0$, $-\sqrt{3}/2$, $-\sqrt{3}/2$, etc.
The values alternates between $-\sqrt{3}/2$, $0$ and $\sqrt{3}/2$. It does not look like it gets closer to any particular value. But we want to find an explicit $\varepsilon$ which does not work. Well, I need to be able to find values at least $2 \varepsilon$ apart. But I can make values $\sqrt{3}$ apart, by choosing $n$ and $m$ such that $a_n = \sqrt{3}/2$ and $a_m = -\sqrt{3}/2$. 
So let us take $2 \varepsilon = \sqrt{3}$, so $\varepsilon = \sqrt{3}/2$.

Now, let us write things down. Assume that the sequence converges to some real number $a$. Fix $\varepsilon = \sqrt{3}/2$. Then there exists an integer $N \geq 0$ such that, for all $n \geq N$,
$|a_n-a| < \sqrt{3}/2.$
But then, for all $n$, $m \geq N$,
$|a_n-a_m| < \sqrt{3}.$
Take $n := 6 (\lfloor N/6 \rfloor+1)+1$ and $m := 6 (\lfloor N/6 \rfloor+1)+4$. Then $n$ and $m$ are both larger than $N$, and $a_n = \sqrt{3}/2$, and $a_m = -\sqrt{3}/2$. Hence,
$|a_n-a_m| = \sqrt{3},$
and we have a contradiction. Hence, $(a_n)$ does not converge.
