By proving $\displaystyle \lim_{n\to\infty}(\cos n)^n \neq 0,$ or any other method you like, show that $\displaystyle \sum_{n=1}^{\infty} (\cos n)^n$ is divergent.
Far from a rigorous proof but seeing as there hasn't been a reply I thought I might as well say something because I thought this limit was quite interesting.
If we first define $\displaystyle p_n \in \mathbb{N} $ such that $\displaystyle p_n = \lfloor {(\pi \times 10^n)} \rfloor $
Then I think the reason the limit doesn't converge to zero is that as $\displaystyle n \to \infty, \quad p_n \to 314159265358979\ldots $ and so $\displaystyle \lim_{n \to \infty} \cos(p_n) = \pm 1$
Like I said, far from rigorous, but nice question anyway.
Thanks
pomp.
How about this:
The set $\displaystyle \{|cos(n)|\},n\in\mathbb{N}^{+}$ is dense in $\displaystyle (0,1)$ and therefore their exist an $\displaystyle \epsilon>0$ such that $\displaystyle 1-|\cos(n)|<\epsilon$ for some $\displaystyle n\in \mathbb{N}^{+}$ and therefore $\displaystyle |\cos(n)|^n >0$ for some $\displaystyle n\in \mathbb{N}^{+}$ no matter how large n.