Since the sequence $\langle a_n\rangle=\langle 1-\cos(1/n)\rangle$ is convergent, so $\limsup a_n=\liminf a_n=\lim a_n =0.$ At first, I calculated the limit superior and the limit inferior of $\langle a_n\rangle$ using the traditional method which is by determining the two sequences: $b_n=\sup\{a_n,a_{n+1},\ldots\}$ and $c_n=\inf\{a_n,a_{n+1},\ldots\}$ followed by taking their $\inf$ and $\sup,$ respectively. As expected, I obtained $\limsup a_n=0=\liminf a_n.$
However, we also know that if $L$ is the set of all limit points of $\langle a_n\rangle,$ then $\limsup a_n=\sup L$ and $\liminf a_n =\inf L$ (which is equivalent to saying that the limit superior is the greatest subsequential limit and the limit inferior is the least subsequential limit). In the present case, $L=[0,1-\cos 1].$ So, $\limsup a_n=1-\cos 1$ and $\liminf a_n=0.$ The results obtained here are absurd. I must be missing something obvious! It seems that $L$ should be the singleton set $\{0\}.$ But, $a_n=1-\cos(\frac1n)=2\sin^2(\frac1{2n})\in(0,2\sin^2(1/2))=(0,1-\cos 1).$ This implies $L=[0,1-\cos 1],$ as I mentioned before.
Please help me see where I went wrong...
Though the sequence starts with $1-\cos 1$, this number has nothing to do with limit points. The first few terms of a sequence have no effect on the limit points. The only limit point of this sequence is $0$.