What are the limit points of set $E=\left\{ a_{n}\::\:n=1,\:2,\:3,\:\cdots\right\} $, $a_{n}=\frac{n-1}{n+1}\left|\sin\frac{n\pi}{7}\right|$

Question

Find the limit points of the following sequence:

Take the sequence:$$a_{n}=\frac{n-1}{n+1}\left|\sin\frac{n\pi}{7}\right|$$

We wish to find: a. the limit points b. $\underline{\lim}_{n\rightarrow\infty}a_{n}$, and $\overline{\lim}_{n\rightarrow\infty}a_{n}$ My attempt: When $\frac{n\pi}{7}=k\pi$, i.e, $n=7k$, $k$ are integers, then we have $\underline{\lim}_{n\rightarrow\infty}a_{n}=0$

When $\frac{n\pi}{7}=\frac{\pi}{2}+k\pi$, i.e, $n=7\left(k+\frac{1}{2}\right)$, $k$ are integers, then we have $\overline{\lim}_{n\rightarrow\infty}a_{n}=1$

Is that correct?

How to find the limit points? Thank you very much

Answer 1

Note that

$$\left\{\left\vert \sin\dfrac{n\pi}{7}\right\vert \ ; \ n \in \mathbb N\right\} \subseteq \{0, \sin \dfrac{\pi}{7},\sin \dfrac{2\pi}{7}, \sin \dfrac{3\pi}{7}\}=A$$

As $\lim\limits_{n \to \infty} \dfrac{n-1}{n+1}= 1$, the set $L$ of limit points of $\{a_n \ ; \ n\in \mathbb N\}$ is included in $A$. By picking appropriate subsequences, you can prove that $L=A$.

Then $\liminf a_n =0$ and $\limsup a_n = \sin \dfrac{3\pi}{7}$.

Answer 2

If $n=7m+k$, $k=0,1,2,3,4,5,6$ then $\left|\sin\frac{(7m+k)\pi}{7}\right|=sin(k\pi/7)$. But also $lim\frac{n-1}{n+1}=1$,so the available limit points are: $sin(k\pi/7)$. What is the biggest and smallest of these values? these are upper and lower limit points you are looking for.