Find lim sup lim inf

Question

How do I find $\lim \sup\text{ or } \lim \inf$ of $ \sin (\frac{n\pi}{5})$ ?

I know the $\sin$ function normally oscillates between $-1$ and $1$ but that obviously is not the answer for $\lim \inf$ and $\lim \sup$.

Answer 1

Observe that $$\begin{align*}&\left\{ \sin \left(\frac{n\pi}{5}\right)|n\in \mathbb{N}\right\}\\&=\left\{\sin \left(\frac{\pi}{5}\right),\sin \left(\frac{2\pi}{5}\right),\sin \left(\frac{3\pi}{5}\right),\sin \left(\frac{4\pi}{5}\right),\sin \left(\frac{5\pi}{5}\right),\sin \left(\frac{6\pi}{5}\right),\sin \left(\frac{7\pi}{5}\right),\sin \left(\frac{8\pi}{5}\right),\sin \left(\frac{9\pi}{5}\right),\sin \left(\frac{10\pi}{5}\right)\right\} \\&=\left\{0,\sin \left(\frac{\pi}{5}\right),\sin \left(\frac{2\pi}{5}\right),\sin \left(\frac{3\pi}{5}\right),\sin \left(\frac{4\pi}{5}\right),-\sin \left(\frac{\pi}{5}\right),-\sin \left(\frac{2\pi}{5}\right),-\sin \left(\frac{3\pi}{5}\right),-\sin \left(\frac{4\pi}{5}\right)\right\}.\end{align*}$$

Therefore

$$\begin{align*}&\limsup_{n\to\infty} \sin \left(\frac{n\pi}{5}\right)\\&=\max \left\{0,\sin \left(\frac{\pi}{5}\right),\sin \left(\frac{2\pi}{5}\right),\sin \left(\frac{3\pi}{5}\right),\sin \left(\frac{4\pi}{5}\right),-\sin \left(\frac{\pi}{5}\right),-\sin \left(\frac{2\pi}{5}\right),-\sin \left(\frac{3\pi}{5}\right),-\sin \left(\frac{4\pi}{5}\right)\right\}\end{align*} $$ and

$$\begin{align*}&\liminf_{n\to\infty} \sin \left(\frac{n\pi}{5}\right)\\&=\min \left\{0,\sin \left(\frac{\pi}{5}\right),\sin \left(\frac{2\pi}{5}\right),\sin \left(\frac{3\pi}{5}\right),\sin \left(\frac{4\pi}{5}\right),-\sin \left(\frac{\pi}{5}\right),-\sin \left(\frac{2\pi}{5}\right),-\sin \left(\frac{3\pi}{5}\right),-\sin \left(\frac{4\pi}{5}\right)\right\}.\end{align*} $$

Answer 2

First, you need to know the definition of $lim sup$ and $lim inf$. Given a bounded sequence $ \sin (\frac{n\pi}{5})$. Let's name it $\{a_j\}$. We define $$A_1=inf\{a_1,a_2,a_3,...\}$$ $$A_2=inf\{a_2,a_3,a_4,...\}$$ $$A_3=inf\{a_3,a_4,a_5,...\}$$ ... $$A_n=inf\{a_n,a_{n+1},...\}$$ It's easy to find the above lines shares the same infimum, because sin function is periodical. But it's not obvious to see which is the infimum (here is also minimum.) Hence, I use R to help me find:

x <- seq(pi/5, 2*pi, by = pi/5)
min(sin(x))

R gives me -0.9510565. Hence, $$lim inf{a_j}=limA_n=-0.9510565$$

Similarly, $$lim sup{a_j}=limB_n=0.9510565$$, where $B_n=sup\{a_n,a_{n+1},...\}$