TITLE: How does one show that $(n-1)!\cos(n!x)$ diverges everywhere?
QUESTION [5 upvotes]: In particular, how does one show divergence at points for which the fractional part of $n!x$ tends to $\frac{1}{2}$ as $n\to\infty$, such as $x=\frac{1}{2}\sum\frac{1}{(2k)!}$?
(This is an oft-cited example of a sequence of functions which are derivatives of everywhere uniformly convergent functions, but which converge nowhere.)

REPLY [3 votes]: How does one show that $(n-1)!\cos(n!x)$ diverges everywhere?

Let $x_n=(n-1)!\cos(n!x)$. Assume that the sequence $(x_n)$ is bounded.
Then $\cos(n!x)\to0$ when $n\to\infty$ hence, for every $n$, 
$$
n!x=\frac\pi2+k_n\pi+z_n,
$$ 
where every $k_n$ is an integer valued and $z_n\to0$. Thus, $|\cos(n!x)|=|\sin(z_n)|\sim |z_n|$ hence $(n-1)!\cdot|z_n|\leqslant z$ for every $n$, for some finite $z$. In particular, 
$$
(2n)!x=(2n)\cdot(2n-1)!x=(n+2nk_{2n-1})\pi+2nz_{2n-1}=2nz_{2n-1}\pmod{\pi}.
$$ 
Since $2nz_{2n-1}\to0$, $|\cos((2n)!x)|\to1$. This implies that $|x_{2n+1}|\to\infty$, which contradicts the hypothesis that the sequence $(x_{n})$ is bounded.
In particular, the sequence $(x_n)$ diverges.
The same reasoning shows that the sequence $(x^{(a)}_n)$ diverges (for being unbounded) for every $a\gt1$, where $x^{(a)}_n=n^a\cos(n!x)$.