Accumulation points of a trignometric sequence

Question

I am looking for the limit points of the sequence $$<f_n>= \sin\left(\frac{n\pi}{2}\right) + \cos\left(\frac{n\pi}{3}\right).$$ I solved and found the sequence as $$<\frac{3}{2},\frac{-1}{2},-2,\frac{-1}{2},\frac{3}{2},1,\frac{-1}{2},-1,...........>$$ However I am unable to find the limit points. Please help.

Answer 1

First, note that $f_{n+12}=f_{n}$ since \begin{align}f_{n+12}&=\sin\left(\dfrac{(n+12)\pi}{2}\right)+\cos\left(\dfrac{(n+12)\pi}{3}\right)\\&=\sin\left(\dfrac{n\pi}{2}+6\pi\right)+\cos\left(\dfrac{n\pi}{3}+4\pi\right)\\&=\sin\left(\dfrac{n\pi}{2}\right)+\cos\left(\dfrac{n\pi}{3}\right)\\&=f_n\end{align} So only the twelve first terms are important. $$\frac32, -\frac12, -2, -\frac12, \frac32, 1, -\frac12, -\frac12, 0, -\frac12, -\frac12, 1$$ To have the limit points, we remove all duplicate from this list. $$-2, -\frac12, 0, 1, \frac32$$ These are the limit points of your sequence.