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!
1 Answer
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 $\lfloor n S_n \rfloor = \lfloor ne \rfloor$.
