Show that $\sin \dfrac{n \pi}{4}$ is divergent. Show that $\sin \dfrac{n \pi}{4}$ is divergent. 
My attempt:
Consider the subsequences 
$x_{4n}=\sin (n \pi)$, which converges to $0$, 
and $x_{8n+2}=\sin \dfrac{2(4n+1) \pi}{4}$, which converges to $1$. 
As the two subsequences converge to two different limits, the sequence is divergent.
My questions: 


*

*Are these two indeed subsequences of the given sequence, specially the second one?

*I found the limits inductively. Is there any analytic way to show it? 
Thanks in advance :) 
 A: A subsequence of a sequence $(f(n))_{n}$ is any sequence of the form $(f\circ g(n))_n$ where $g:\Bbb N \to \Bbb N$ is an increasing function. In your case, the defined sequences are indeed subsequences of $(x_n)_n$ taking $g(n)=4n$ and $g(n)=8n+2$ in each case.
Now, for each $n\in \Bbb N$:$$x_{4n}=\sin(n\pi)=0$$
and 
$$x_{8n+2}=\sin \left(\frac{(8n+2)\pi}{4}\right)=\sin \left((4n+1)\frac{\pi}{2}\right)=1$$
so not only the limits are zero and one respectively, but each element of the sequence equal $0$ and $1$ respectively.. What do you mean with inductively?
A: For question 1.: Yes, and for question 2: Consider $a_n = \sin \left(\frac{n\pi}{4}\right)$, and look at $\lim \text{ sup $a_n$}$ and $\lim \text{ inf $a_n$}$ and show they are different. You can infact determine these limits.
A: You can certainly take every $x_{8n+2}$ for consecutive positive integers $n$, and it is a subsequence of $x_n$. Your arguments look fine.
It is a general fact that $\sin(2n\pi + \alpha) = \sin(\alpha)$
for any integer $n$ and real number $\alpha$.
In the case of your first sequence, by this fact
all elements are equal either to $\sin 0$ or to $\sin \pi.$
For the second sequence, the same fact shows that all elements are
equal to $\sin(\frac\pi2).$
I would probably prefer induction to prove that
$\sin(2n\pi + \alpha) = \sin(\alpha)$,
but you only need to do that proof once and you can then apply it
to many problems like this one.
So in that sense, this particular problem does not require induction
(at least in my opinion).
