The book I am using for my Advance Calculus course is Introduction to Analysis by Arthur Mattuck.
Find the cluster points of:
(a). {${sin(\frac{n+1}{n} \frac{n}{2})}$}
(b). {${sin(n+\frac{n}{2})\frac{\pi}{2}}$}
For each cluster point, find a sub-sequence converging to it.
This is my rough proof to this question. I was wondering if anybody can look over it and see if I made a mistake or if there is a simpler way of doing this problem. I want to thank you ahead of time it is greatly appreciated.So lets begin.
Proof:
I think you understand the problem, but let me write the answer in a more practical way.
(a) Your sequence is $ \sin \big( \frac{n+1}{n } \frac{\pi}{2} \big) $ is covergent, the set of the cluster point is $ \{ \sin \big( \frac{n+1}{n } \frac{\pi}{2} \big) , n \in \mathbb{N}\} \cup \{ 1\}$. To find a sunsequence for each one is an easy task.
(b) In this case, your sequence is $ \sin \big( \frac{3n}{2 } \frac{\pi}{2} \big) $, This sequence has three subsequential limits $ \{ -1,0, \frac{\sqrt{2}}{2} \}$, hence it is divergent. The set of cluster points is $ \{ \sin \big( \frac{3n+}{2n } \frac{\pi}{2} \big) , n \in \mathbb{N}\} \cup \{ -1,0, \frac{\sqrt{2}}{2} \}$ . To find the subsequences of these cluter points you may take $ n= 4k, \; 4k+1, \; 4k+2$ respectively.