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I was not allowed to edit the previous question because people spent their time and energy constructing an answer, therefore I have asked it again, with my initial error in formulating my answers corrected.

Is it true that this example I came up with, namely the sequence $$a_n=\cos\left(\frac{\pi}{3} + \frac{\pi n}{2}\right)$$ Has exactly $4$ limit points/accumulation points? What I tried to do was symmetrically pick an infinite amount of points that lie on the unit circle and all hop between four exact points $\pm \frac{1}{2}$ and $\pm \frac{1}{2} \sqrt{3}$.

Is equality fine for accumulation points is basically what I am asking.

I also was thinking of an example with infinitely many accumulation points and wondered if the same example would cut it:

$$b_n = n \cos(n) $$ Since it oscillates back and forth it comes back to every single point eventually.

As an alternative way of getting four accumulation points I could define unions of the sets $s_a= \{ a+ \frac{1}{n}| n \in \mathbb{N} \}$ for values $a=0, 1,2,3$ but this example feels very contrived and I do not know how to formulate this in terms of a sequence other than just listing the elements one by one in set notation: $\{ \frac{1}{1}, 1+\frac{1}{1}, 2+\frac{1}{1} \dots, 1+ \frac{1}{n}, 2+ \frac{1}{n} \dots \}$

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    You have asked a question with the same title less than one hour ago, moreover without mentionning it. This is not a good practice. You should have modified the text of the initial question instead of asking a new one. – Jean Marie Nov 23 '18 at 23:00

1 Answers1

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We have that

  • $n=1 \implies a_1=\cos\left(\frac{\pi}{3} + \frac{\pi }{2}\right)=-\frac{\sqrt 3}2$
  • $n=2 \implies a_2=\cos\left(\frac{\pi}{3} + \pi\right)=-\frac12$
  • $n=3 \implies a_3=\cos\left(\frac{\pi}{3} + \frac{3\pi }{2}\right)=\frac{\sqrt 3}2$
  • $n=4 \implies a_4=\cos\left(\frac{\pi}{3} + 2\pi\right)=\frac12$
  • $n=5 \implies a_5=a_1$

For the second question, yes of curse $b_n$ oscillates diverging assuming values $\in \mathbb{R}$ and $\cos n$ is dense in $[-1,1]$. From that to conlcude that $n\cos n$ is dense on the real line is not so trivial, see the related

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