Is $\{n!\alpha\},n \in\mathbb{N}$ dense in $R$ ? where $\alpha$ is irrational Is $\{n!\alpha\},n \in\mathbb{N}$ dense in $[0,1]$ ? where $\alpha$ is irrational.
I know that $\{n\alpha\}$ is dense in $[0,1]$? I wanted to generalise it.So, first I thought about $\{n!\alpha\},n \in\mathbb{N}$, but couldn't make any reasonable progress, although my intuition says answer will be no.

You can work out, through a somewhat careful argument, that if a sequence $s_n$ increases without bound, then $\{s_n\alpha\}$ is dense in $[0,1]$ for almost every real $\alpha$. However, in general, this sequence is not dense for every irrational $\alpha$.
We can actually find a particular notable counterexample for $n!$. Consider
$$e=\sum_{k=0}^{\infty}\frac{1}{k!}.$$
Note that $$\{n!e\} = \sum_{k={n+1}}^{\infty}\frac{n!}{k!}$$
since the term up to the $n^{th}$ are integers after multiplication with $n!$ and the remaining terms (assuming $n > 1$) do not sum up to $1$. In fact, note that $\frac{n!}{k!} \leq \left(\frac{1}{n}\right)^{k-n}$ which implies, using a geometric sum, that 
$$\{n!e\} = \sum_{k={n+1}}^{\infty}\frac{n!}{k!} \leq \sum_{i=1}^{\infty}\left(\frac{1}n\right)^i = \frac{1}{n-1}.$$
Of course, a sequence satisfying this cannot possibly be dense. 
You might notice that the fact that we have $n!$ as opposed to some other similar sequence is not that important; it's possible to extend this to give counterexamples for any chosen sequence $s_n$ of integers where each term divides the next - but it's a bit nice that the counterexample here is a well-known number.
