Prove that $(y_n)$ converges Prove that $(y_n) = \frac1n\sin({n\pi\over3})$ converges
Now I know my RTP:  ($\forall\epsilon\gt0)(\exists k \in N)(\forall n \gt k) \\ |(y_n)-c| \lt \epsilon $
but from there i get stuck.
 A: We show $(y_n)\rightarrow 0$. Let $\varepsilon>0$ be given. Choose $N>\frac{1}{\varepsilon}$, and suppose that $n\geq N$. Note that $\left\lvert \sin\left(\frac{n\pi}{3}\right)\right\lvert\leq 1$, so
\begin{align}
\left\lvert \frac{1}{n}\sin\left(\frac{n\pi}{3}\right)\right\lvert \leq \frac{1}{n}<\varepsilon.
\end{align}
A: notice sine function is always bounded by $1$ for every number in its argument. So, the absolute value of your sequence if bounded. In particular, the absolute value of all the $y_n's$ are less equal than $\frac{1}{n}$ that is easy to deal with. In particular, take your $N$ to be $1/ \epsilon $ in the $\epsilon - N$ definiton of the limit of a sequence.
A: Hint: Use squeeze theorem.

Let $(a_n), (b_n)$ be infinity sequence, such that $a_n =\frac{1}{n} \cdot (-1)
\wedge b_n = \frac{1}{n} \cdot 1$. Now just use above theorem.
$$\left(
\left(\forall n \in \mathbb {N}^+\right)\left(a_n \leq y_n \leq c_n\right) \wedge \left(\lim_{n\to\infty}a_n=0 = \lim_{n\to\infty}b_n\right) 
\right)
\Longrightarrow
\lim_{n\to\infty}y_n=0$$
