# Limit superior and limit inferior of $\{cos\frac{n \pi}{2} \}_{k=1}^\infty$

I'm trying to find the limit superior and limit inferior of $\{cos\frac{n \pi}{2} \}_{k=1}^\infty$.

The definitions that I'm using are

$\displaystyle \limsup\{a_n\} = \lim_{n \rightarrow \infty} [sup \{a_k: k \geq n\}]$

$\displaystyle \liminf\{a_n\} = \lim_{n \rightarrow \infty} [inf \{a_k: k \geq n\}]$.

I've done some research and it looks like you can find the limit superior and limit inferior by taking the supremum and infimum over the collection of all possible subsequential limits.

So for my problem, if we write out some of the terms,

$\{cos\frac{n \pi}{2} \}_{k=1}^\infty = \{0, -1, 0, 1, .....\}$ and it repeats like this.

I have four sub sequential limits $0, -1, 1$.

So $\limsup\{cos\frac{n \pi}{2} \} = 1$

and $\liminf\{cos\frac{n \pi}{2} \} = -1$

Is my approach correct?

Thanks.

• The moment you know well the definitions of both corresponding limits you have almost the evidence that your answers are correct. (I say your answers, not your methods I did not saw). – Piquito Feb 13 '17 at 0:30

Yes. In fact, the limit superior of a sequence $(x_n)$ is precisely the supremum of the set $S$ of subsequencial limits of $(x_n)$, and analogously, the limit inferior is the infimum of $S$.
Also, there will always be a subsequence of $(x_n)$ converging to its limit superior, and likewise there will always be a subsequence of $(x_n)$ converging to its limit inferior. That is to say, $\limsup x_n, \, \liminf x_n \in S$.
In your example where $x_n = \cos \left(\frac{n\pi}{2} \right)$, you found a subsequence converging to $1$, and this sequence $x_n$ is never any bigger than $1$ by the nature of cosine, so $1$ is automatically the largest a subsequencial limit could be.