$\lim_{n\to \infty} n!e - \lfloor n!e \rfloor=?$ I want to compute the limit in the title, I know that the term in the limit is bounded from below by zero, and I want somehow to bound it from above by a term that converges to zero;
I thought of using the sum definition of $e=\sum_{k=0}^\infty 1/k!$, but how exactly?
 A: Note that $$n!e=\sum_{k=0}^n\frac{n!}{k!}+\sum_{k\ge n+1}\frac{1}{\prod_{j=n+1}^{k}j}.$$The first sum is of integers, while the second is of positive fractions bounded above by a geometric progression $$\sum_{k\ge n+1}\frac{1}{(n+1)^{k-n}}=\frac{1}{n}.$$In fact this is an overestimate, so even for $n=1$ the terms total $\lt1$. Thus $\lfloor n!e\rfloor$ is the sum of integers we put to one side, whereas $$n!e-\lfloor n!e\rfloor=\sum_{k\ge n+1}\frac{1}{\prod_{j=n+1}^{k}j}.$$
A: Render
$e=\sum_{k=0}^{\infty}(1/k!)$
From this
$n!e=\sum_{k=0}^{\infty}(n!/k!)$
where the summation term is an integer whenever $n\ge k$.  Then show that the given difference in the problem will be
$\sum_{k=n+1}^{\infty}(n!/k!)=(1/(n+1))+(1/((n+1)(n+2)))+...$
Render that difference less than
$\sum_{k=n+1}^{\infty}(1/(n+1)^{k-n})=(1/((n+1)+(1/(n+1)(n+1)))+...$
and use the formula for the sum of a geometric series to evaluate this comparison series.  Once you do that the squeeze theorem will, ahem, zero in on the answer.
