<s>[INST] On convergence of $\cos(2\pi \alpha n!)$ 
Let $x\in \mathbb R$
Prove that there exists some $\alpha_x$ such that the sequence $\cos(2\pi \alpha_x n!)$ converges to $\cos(x)$

I've been stumped with this problem for a while. I've been looking for some $\alpha$ such that $$\forall n, 2\pi \alpha n! = x + 2\pi k_n + \epsilon_n$$
where $k_n$ is an integer and $\epsilon_n$ a sequence that goes to $0$.
Can someone help me build the $\alpha$ ?
 [/INST] For any $x \in \mathbb{R}$, take a $y \in [0,\pi]$ such that $\cos x = \cos y$.
Define
$$\alpha = \sum_{n=1}^\infty \frac{1}{n!}\left\lfloor \frac{y}{2\pi} n\right\rfloor$$
For any $N > 0$, we have
$$\alpha N! = 
\underbrace{\sum_{n=1}^N \frac{N!}{n!}\left\lfloor \frac{y}{2\pi} n\right\rfloor}_{\in \mathbb{Z}}
+ \frac{1}{N+1}
\left\lfloor \frac{y}{2\pi} (N+1)\right\rfloor
+ \sum_{k=2}^\infty \frac{1}{\prod_{j=1}^{k-1} (N+j)}\left(\frac{1}{N+k}\left\lfloor \frac{y}{2\pi}(N+k)\right\rfloor\right)
$$
For the $2^{nd}$ term, it is a number near $\frac{y}{2\pi}$. More precisely,
$$0 \le \frac{y}{2\pi} - \frac{1}{N+1}\left\lfloor \frac{y}{2\pi} (N+1)\right\rfloor < \frac{1}{N+1} < \frac{1}{N}$$
For the $3^{rd}$ term, we have the bound
$$0 \le \sum_{k=2}^\infty \frac{1}{\prod_{j=1}^{k-1} (N+j)}\left(\frac{1}{N+k}\left\lfloor \frac{y}{2\pi}(N+k)\right\rfloor\right)
\le \sum_{k=2}^\infty \frac{1}{(N+1)^{k-1}}\frac{y}{2\pi}
= \frac{y}{2\pi N} < \frac{1}{2N}
$$
Combine these two bounds, we can conclude
$$\alpha N! = \beta_N + \frac{y}{2\pi} + \delta_N
\quad\text{ where }\quad \beta_N \in \mathbb{Z}\;\text{ and }\; -\frac{1}{N} < \delta_N < \frac{1}{2N}$$
Apply MVT to $\cos(t)$ for $t$ between $2\pi\alpha N!$ and $2\pi\beta_N + y$, we find
$$|\cos(2\pi\alpha N!) - \cos(y)| \le 2\pi|\delta_N| < \frac{2\pi}{N}$$
As a result,
$$\lim_{N\to\infty} \cos(2\pi\alpha N!) = \cos y = \cos x$$
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