Show that $\lim\limits_{n\to\infty}n\cos(n)$ is divergent by definition of limit. I am trying to split it into two cases, (1) $\cos(n)\geq 0$ and (2) $\cos(n)<0$. Then for (1), I would like to show the limit diverges to $+\infty$, and $-\infty$ for (2).
Then I tried to formulate the definition of limit diverging to $\pm\infty$, by using the Archimedean property, but it seems not working. Here are the details:
Say for (1): $n\in\left[ 2k\pi,\pi/4+2k\pi\right]\cup\left[ 3\pi/4+2k\pi, 2\pi+2k\pi\right]$ , and I want to get something like
$$\forall M\in \mathbb{R},\exists K\in \mathbb{N} \text{ s.t. } n\cos(n) >M \text{ for every } n\geq K$$
Then by Archimedean property,
$$ \forall M\in \mathbb{R}, \exists K=[N\cos(N)]\text{ s.t. } K>M$$
But $\cos(n)$ actually depends on $n$. Say for $n\geq K$, when $\cos(n)$ is very small, $n\cos(n)$ may not be greater than $M$.  I am wondering if there is any other possible way to tackle this problem.
Or is there any way to find two subsequences that converging to different limits? Since $n\in \mathbb{N}$, I find it kinda subtle to deal with cosine.
 A: You can prove this by proving that a subsequence cannot have a stationary point.
Use
$$\cos(2x)=2\cos^2(x)-1$$
Simply speaking, a series created out of this equation would have to have a limit $0$ if the asked limit is to exist, but this series cannot have $0$ as a limit.
So, based on the above equation create a series
$$u_{n+1}=2u_{n}^2-1, u_0=\cos(1)$$
Now to get $g_n=2^n\cos(2^n)$ which is a subsequence that we are going to track simply multiply everything by $2^{n+1}$
$$2^{n+1}u_{n+1}=2^{n+2}u_{n}^2-2^{n+1}$$
or
$$g_{n+1}=\frac{g_n^2}{2^{n-2}}-2^{n+1}$$
Even from here it is obvious that limit does not exist, but let us elaborate.
$$\frac{g_{n+1}}{2^{n+1}}=\frac{g_n^2}{2^{2n-1}}-1$$
Replace:
$$t_n=\frac{g_{n}}{2^{n}}$$
$$t_{n+1}=2t_{n}^2-1$$
If there is a limit for $t_n$ it is a fixed point (a solution) for
$$x=2x^2-1$$
And this has a solution, none of which is $0$. But then
$$g_n=2^{n}t_n$$
cannot have a limit as it becomes unbounded. Notice that $t_n$ must tend to $0$ if $g_n$ is to have a limit, but $0$ is not even a stationary point for $t_n$.
$\cos(n)$ not having a stationary point, but still being bounded, is implying that it is not affecting the nonexistence of a limit of any unbounded increasing function like $n$. For this conclusion, we do not need to know the distribution of the values of $\cos(n)$.
A: Definition of the limit of a sequence in $\mathbb R$
$$
(\{ a_k\}_{k \in \mathbb N} \text{ converges to }a  \in \mathbb R ) \iff \exists a \in \mathbb R :\forall \varepsilon > 0: \exists N\in  \mathbb N: \forall n > N: |a_n - a| < \varepsilon
$$
Negation of the definition
$$
(\{ a_k\}_{k \in \mathbb N} \text{ diverges in } \mathbb R ) \iff \forall a \in \mathbb R :\exists\varepsilon > 0: \forall N\in  \mathbb N: \exists n > N: |a_n - a| \ge \varepsilon
$$
To show divergence: Since every convergent sequence is Cauchy then just show that the given sequence in not Cauchy
A: I'm not so sure about what you say you want or your application of the Archimedean property, but here is my idea.
Note that, because $1<\pi$, the sequence $n\cos(n)$ moves all the time from positive to negative values: in particular, if $m \pi < n < (m+1) \pi$, then either $n+1$ or $n+2$ or $n+3$ or $n+4$ is in the interval $((m+1) \pi, (m+2) \pi)$. It follows that, to prove that your sequence converges, it is enough to show that it doesn't converge to $0$.
We do this by contradiction. Suppose that it does, then there exists $N\in\mathbb{N}$ such that, for every $n\geq N$,
$$
|n\cos(n)|<1 \Rightarrow |\cos(n)|<\frac{1}{n}.
$$
We can take $n$ to be large enough so that $\frac{1}{n}<\frac{1}{\sqrt{2}}$. In this case the above inequality will only hold when $\cos(n)$ is close enough modulo $\pi$ to $\frac{\pi}{2}$, in particular $\cos(n)$ must belong in some interval
$$
(\frac{\pi}{4}+ \pi m, \frac{3\pi}{4}+\pi m) 
$$
for some $m$. But notice that, since $1<\frac{5\pi}{4}-\frac{3\pi}{4}=\frac{pi}{2}$, you will always be able to find $n$ large enough that does not satisfy this. Contradiction.
