$\limsup \cos(\frac{n\pi}{60})$

Question

How do you determine the $\limsup$ / $\liminf$ of $(x_n)_n=\cos(\frac{n\pi}{60})$?

$(x_n)_n$ is bounded, so if I can find the biggest and smallest adherent points I can determine the $\limsup$ and $\liminf$ , but I don't know how...

Answer 1

Let $A$ be the set of cluster points of $(x_n)$. Then $A$ is bounded above by $1$ and below by $-1$.

As $(x_{120k})$ converges to $1$, we have $1\in A$. So $\limsup x_n = \sup A = 1$.

Similarly $(x_{120k+60})$ converges to $-1$ , so $-1\in A$ and $\liminf x_n = \inf A = -1$.

Answer 2

As $\forall n\in N \;\;-1\leq x_n\leq 1$,

and

$\lim_{n\to\infty}x_{120n}=1$

$\lim_{n\to\infty}x_{120n+60}=-1$

you conclude.

Answer 3

Hint: For all n belonging to N {cn}= sup{cos(kπ/60) for all k>=n} The sequence cn is a constant sequence of 1s