Floor function inequality of multiplication

Question

In a final step of a homework, I want to deduce that $$n\lfloor(n-1)!e\rfloor+2\le \lfloor n!e\rfloor+1$$ I'm unable to see whether this is true in general that $$n\lfloor a\rfloor+1\le \lfloor na\rfloor$$ where $n$ is a natural number or do I need to use power series of $e$ to do some reasoning. Thank you.

Answer 1

It is not in general true that $n\lfloor a\rfloor+1\le \lfloor na\rfloor$. In particular, if $n,a$ are both naturals, the floor will not matter and you would be asking for $na+1 \le na$ which is false. So yes, you need the power series of $e$.

Answer 2

You need only one false value of n to invalidate the inequation. Let's try with the first natural number n=1 :

    From n⌊(n−1)!e⌋+2 ≤ ⌊n!e⌋+1 
    we get ⌊(0)!e⌋+2 ≤ ⌊0!e⌋+1 
    and ⌊e⌋+2 ≤ ⌊e⌋+1. 

Wich is obviously false.