How to prove lim sup $\exp\left(-\cos n\right)$ Let $a_n=\text{exp}(-\cos(n))$. I have found that lim sup $a_n=e$ and lim inf $a_n=\frac{1}{e}$. But I'm just starting out with these concepts and I'm not sure how to justify these answers without the usual "cos is bounded above by $1$, below by $-1$ and exp is increasing". Any help on how to be more rigorous here?
 A: This is not an answer yet, as we still need to justify the last 'why' or use another approximation theorem at once. I am not entirely sure if the irrationality of $\pi$ is enough. ( Maybe we should use Weyl's distribution theorem? )
Your sequence given by $a_n= e^{-\cos(n)}$ satisfies $e^{-1}<a_n< e$ for all $n\geq1$ , so to prove that $\limsup a_n= e$ and $\liminf a_n= e^{-1}$ it suffices to prove that :
$\forall \delta>0\ , \forall N\geq 1 : \exists\ n_1, n_2\geq N$ such that $e- a_{n_1}<\delta$ and $a_{n_2}- e^{-1}<\delta \quad $   ( check why? )
To do so, let $\delta>0\ $ and $ N\in \mathbb{N} $ be arbitraries, we are looking for some $n_1\geq N$ such that $e-a_{n_1}= e-e^{-\cos(n_1)}<\delta$ , i.e. $\ e- \delta< e^{-\cos(n_1)}$ or $\ \cos(n_1)< -\ln(e-\delta)$ 
( here the assumption is that $\delta<< e$ otherwise it is verified obviously )
When $\delta\to 0^+$ clearly $\epsilon= -\ln(e-\delta)\to -1^+$ and we are looking for a natural number $n_1$ for which $-1\leq\cos(n_1)< \epsilon$ , so we are going to prove that $\forall \alpha>0 :\ \exists\ n_1>N $ such that $n_1\in (\pi+ 2k\pi- \alpha,\pi+ 2k\pi+\alpha)$ for some $k\in \mathbb{N}$ and we are done. ( why ? )
Toward a contradiction suppose that the above claim is false, i.e. $\exists\ \alpha_o>0 :\ \forall n>N : n\notin (\pi+ 2k\pi- \alpha_o,\pi+ 2k\pi+\alpha_o)$ for all $k\in \mathbb{N}$ , then for all $n>N\ \forall k\in \mathbb{N}:\ |n- (2k+1)\pi|\geq \alpha_o$ , that is $\ |\pi- \frac{n}{2k+1}|\geq \frac{\alpha_o}{2k+1} $
Now recall Dirichlet's approximation theorem; approximate irrational numbers by rational numbers . There are infinitely many pairs $(n,q)$ such that $|\pi- \frac{n}{q}|<\frac{1}{q^2}$ , in particular there are infinitely many such that $n>N$ (why?) and we can choose $q$ to be an odd number ( $\underline{\text{why ?}}$ ) such that $\frac{1}{q^2}< \frac{\alpha_o}{q}$ by choosing a big enough $q$ verifying $\frac{1}{q}<\alpha_o$. This contradicts the hypothesis.
A similar treatment permits to find the other natural $n_2$ for the $\liminf$ , if we can find $q$ to be even in that case!
But how can we make sure that $q$ can be found to be an odd and an even number? As far as Dirichlet's theorem allows they can all be evens or all odds !
A: Really, you just want to know whether or not: $$\limsup_{n\to\infty}\cos(n)=1,\,\liminf_{n\to\infty}\cos(n)=-1$$Is actually true. There are maybe more elementary ways to see this, but if you like topology (which I very much do!) there is an 'elementary' argument using some theory of topological dynamical systems.
Consider the unit circle $\Bbb T=\{z\in\Bbb C:|z|=1\}$ and fix some $\zeta\in\Bbb T$. Consider the continuous 'dynamic' $\varphi:\Bbb T\to\Bbb T,\,z\mapsto\zeta\cdot z$. The 'topological dynamical system' $(\Bbb T;\varphi)$ is a group rotation system: for such systems, surprisingly, every point is (uniformly) recurrent!
Then, the (uniform) recurrence means in particular that every neighbourhood $U$ of $1$ has $\zeta^n\in U$ for infinitely many $n$. What does that really mean? Let's consider neighbourhoods of the form: $$U_\delta=\{z\in\Bbb T:1-\Re z<\delta\}$$For $\delta>0$. If $\zeta^n\in U_\delta$, that means $\cos(2\pi\cdot n\alpha)=\Re\exp(2\pi i\cdot n\alpha)\in(1-\delta,1]$, and this occurs for infinitely many $n\in\Bbb N$.
We can consider $\alpha=\frac{1}{2\pi}\in[0,1)$. Then, the above means: for all $\delta>0$, there exists (infinitely many) $n\in\Bbb N$ such that $|1-\cos(n)|<\delta$. Since $\cos(n)<1$ for all $n$, this implies: $$\limsup_{n\to\infty}\cos(n)=1$$
Since $\alpha$ is irrational, the system is transitive. That means that, for any neighbourhood of $-1$, I can get in there from the starting position $1$. Then, considering (uniform) recurrence at the element $-1\in\Bbb T$ shows that: $$\liminf_{n\to\infty}\cos(n)=-1$$
Considering recurrence at any element $z=\exp(i\theta),\theta\in[0,2\pi)$ (along with transitivity), we find: $$\lim_{k\to\infty}\cos(n_k)=\cos(\theta)$$Along some infinite sequence $(n_k)_{k=1}^\infty$. The same shall apply to $\sin$, too. In particular, the sequence $(\cos(n))_{n=1}^\infty$ accumulates at every point in $[-1,1]$.
What I like about this is that it comes more or less for free as a corollary of some minor topological trickery. It also gives uniform recurrence, which I did not elaborate on, but all that really means is that, in the sequence $n_k$, we may take the distances $n_{k+1}-n_k$ to be bounded. Surprisingly, I don't actually need to know how to approximate $\pi$ to know this! I refer the reader to the first chapters of this book for more detail.
