What is the superior limit of this sequence?

Question

So I've been given this sequence: $a_n:=\cos\left(\frac{2n^2+1}{3n}\pi\right)$ and we are asked to determine its superior limit. My first thought was that I can probably write it (due to continuity of cosine) like this$$\limsup\limits_{n\to\infty} \cos \left(\frac{2n^2+1}{3n}\pi\right)=\cos\left(\limsup\limits_{n\to\infty} \frac{2n^2+1}{3n}\pi\right).$$Is this even right? My second thought was that cosine is biggest at $k\pi;k\in\mathbb{N}$ so i have to find a sequence $(n_k)_{k\in\mathbb{N}}$ so that $a_{n_k}$ tends to go to $k\pi$. Are any of my thoughts valid? How do I make sure that i find the right $n_k$ without it taking too long? I also tried $n_k$ as the sequences of even and uneven numbers but that didn't work. Help would be much appreciated.

Answer 1

For $n = 3k$ we have: $$ a_n = \cos\left(\frac{2n^2 + 1}{3n}\pi\right) = \cos\left(\frac{2n}{3}\pi + \frac{1}{3n}\pi\right) = \cos\left(2k\pi + \frac{1}{9k}\pi\right) = \cos\left(\frac{\pi}{9k}\right) $$ So as $k \to +\infty$ we have $\cos\left(\frac{\pi}{9k}\right) \to 1$, hence limit superior is $1$.

Answer 2

You already have an answer to your question. Concerning your approach based upon$$\limsup\cos(a_n)=\cos\left(\limsup a_n\right),$$this is not true in general, even if you are dealing with a continuous function. If, say $a_n=(-1)^n$ and $f(x)=e^{-x}$, then $\limsup a_n=1$, but $\limsup e^{a_n}=e\ne e^{-1}$.