I checked several thousand natural numbers and observed that $\lfloor n!/e\rfloor$ seems to always be an even number. Is it indeed true for all $n\in\mathbb N$? How can we prove it?

Are there any positive irrational numbers $a\ne e$ such that $\lfloor n!/a\rfloor$ is even for all $n\in\mathbb N$?

**Answer**

Note that:

$$e^{-1}=\sum_{k=0}^\infty \frac{(-1)^k}{k!}$$

Then:

$$\frac{n!}e=n!e^{-1} = \left(\sum_{k=0}^{n} (-1)^k\frac{n!}{k!}\right) + \sum_{k=n+1}^{\infty} (-1)^{k}\frac{n!}{k!}$$

Show that if $a_n=\sum_{k=n+1}^{\infty} (-1)^{k}\frac{n!}{k!}$ then $0<|a_{n}|<1$ and $a_n>0$ if and only if $n$ is odd.

So the when $n$ is odd, the value is:

$$\left\lfloor\frac{n!}{e}\right\rfloor=\sum_{k=0}^{n} (-1)^k\frac{n!}{k!}\tag{1}$$

When $n$ is even it is one less:

$$\left\lfloor\frac{n!}{e}\right\rfloor=-1+\sum_{k=0}^{n} (-1)^k\frac{n!}{k!}\tag{2}$$

Now, almost all of these terms are even. The last term $n!/n!=1$ is odd. When $n$ is odd, the second-to-last term $n!/(n-1)!$ is odd, also. But all other terms are even.

So for $n$ odd, there are two odd terms in the sum, $k=n,n-1$.

For $n$ even, there are two odd terms in the sum, $-1$ and $k=n.$

The trick, then, is to show that the $a_n$ has these properties:

$$\begin{align}

&0<|a_n|<1\\

&a_n>0\iff n\text{ is odd}

\end{align}$$

To show these, we note that $\frac{n!}{k!}$ is strictly decreasing for $k>n$ and $(-1)^k\frac{n!}{k!}$ is alternating. In general, any alternating sum of a decreasing series converges to a value strictly between $0$ and the first term of the sequence, which in this case is $\frac{(-1)^{n+1}}{n+1}.$

**Attribution***Source : Link , Question Author : july , Answer Author : Thomas Andrews*