Question: what are limit points of sequence $<f_n>=<\sin{\frac{nπ}{2}}+\cos{\frac{nπ}{3}}>$?
Note: by limit point of sequence, I mean the accumulation point of sequence!
My attempt: I can see, $3/2, -1/2, -2, 0,1$ repeats infinitely many times in the sequence and hence they must be limit point of sequence (however I don't know, is there is any other limit point?) further as range set of this sequence is finite and hence there is no limit point of range set of this sequence, so there will be no limit point of sequence, which is also a limit point of range set of sequence.
But, to calculate these limit points I need to evaluate trigonometric terms in sequence (upto 10 to 12 terms of sequence!). This takes too much time.
Is there is any other way to do this?
I can see given sequence is sum of two bounded sequences, $<g_n>=<\sin{\frac{nπ}{2}}>$ and $<h_n>=<\cos{\frac{nπ}{3}}>$. Further it is easy to calculate limit points of $<g_n>$ and $<h_n>$. Limit points of $<g_n>$ are $-1,0,1$ and limit points of $<h_n>$ are $\frac{1}{2}, \frac{-1}{2}, 1, -1$.
So is from above information (i.e. from limit points of $<g_n>$ and $<h_n>$) can we determine limit points of $<f_n>$? Or is there is any other way?
Please help me,...
$f_n$ is periodic with period $12$ because when you add $12$ to $n$ you add $6\pi$ to the argument of the $\sin$ function and $4\pi$ to the argument of the $\cos$ function. Compute $f_n$ for $12$ values in succession, like $0$ to $11$. All those values are accumulation points and no others.