Equality of Floors of some Partial Sums

Question

Let $S_n=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots+\frac{1}{n!}$ denote the $(n+1)^{st}$ partial sum in the series expansion for $e=\sum_{k\ge 0}\frac{1}{k!}$. I want to prove that $\lfloor n\cdot(S_n+1/n!)\rfloor=\lfloor n\cdot S_n\rfloor$ for $n\ge 3$. That is, $$\left\lfloor\frac{n}{0!}+\frac{n}{1!}+\frac{n}{2!}+\cdots+\frac{n}{n!}\right\rfloor=\left\lfloor\frac{n}{0!}+\frac{n}{1!}+\frac{n}{2!}+\cdots+\frac{n}{n!}+\frac{n}{n!}\right\rfloor.$$ I've tried to write out some kind of induction argument and use the fact that $\lfloor x+y\rfloor\ge\lfloor x\rfloor +\lfloor y\rfloor$, but I just get really messy expressions that don't seem to lead anywhere. I'm not even sure if induction is the right approach here. Any help is greatly appreciated, thanks in advance!

Answer 1

Assume $n \geq 3$.

$S_n$ is a rational number and in lowest terms its denominator is $n!$. Therefore $nS_n$ has lowest terms denominator $(n-1)!$, so $\lceil nS_n \rceil - n S_n \geq \frac{1}{(n-1)!}$. If equality held in this expression then we would have $nS_n + \frac{1}{(n-1)!} = m$ for integer $m$, so $n! S_n + 1 = (n-1)!m$. But $n! S_n$ and $(n-1)!$ are both even numbers, a contradiction. Therefore $\lceil nS_n \rceil - n S_n >\frac{1}{(n-1)!}$.

Since $nS_n$ and $n(S_n + \frac{1}{n!})$ differ by $\frac{1}{(n-1)!}$, this implies that $\lfloor n S_n \rfloor = \lfloor n(S_n + \frac{1}{n!}) \rfloor$. QED

Side note: Using the remainder of the Taylor series for $e^x$ at $x=1$, $0 < e - S_n \leq \frac{e}{(n+1)!}$ (Lagrange remainder). This easily implies that $\lfloor n S_n \rfloor = \lfloor ne \rfloor$.