Prove that $(y_n)$ converges

Question

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.

Answer 1

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}

Answer 2

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.

Answer 3

Hint: Use squeeze theorem.

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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.

$$\left( \left(\forall n \in \mathbb {N}^+\right)\left(a_n \leq y_n \leq c_n\right) \wedge \left(\lim_{n\to\infty}a_n=0 = \lim_{n\to\infty}b_n\right) \right) \Longrightarrow \lim_{n\to\infty}y_n=0$$