Today I read about rencontres number, but I don't exactly understand some part of the proof of the formula. Can someone explain?
One of those is the limit $$ !n=n(!(n-1))+(-1)^n \, \Rightarrow \lim_{n\to\infty} \frac{!n}{n!} =\frac{1}{e} $$ The other thing is how the formula is proved for $n$ where n is the amount of units in the sequence, but when we apply it, we use $[\frac{(n-k)!}{e}]$.
I tried googling the explanation, but I haven't found anything simple enough for myself to understand.
For the limit, notice
$$!n=n(!(n-1))+(-1)^n$$
$$=n((n-1)(!(n-2))+(-1)^{n-1})+(-1)^n$$
$$=\ldots$$
$$=[1](-1)^n + [n](-1)^{n-1} + [n(n-1)](-1)^{n-2} + \ldots + [n(n-1)\cdots 1](-1)^{0}$$
$$=\Sigma_{i=0}^n [\frac{n!}{i!}](-1)^i$$
$$=n!\Sigma_{i=0}^n \frac{(-1)^i}{i!}$$
And the part after $n!$ is (taylor series) definition of $\frac{1}{e}$
In fact, you can carry on after getting the series representation of $!n$.
$$!n=n!\Sigma_{i=0}^n \frac{(-1)^i}{i!}$$
$$|\frac{n!}{e}-!n|=|\Sigma_{i=n+1}^\infty(-1)^i\frac{n!}{i!}|$$
$$=|\frac{1}{n+1}-(\frac{1}{(n+1)(n+2)} - \frac{1}{(n+1)(n+2)(n+3)}) - \cdots|$$
$$< \frac{1}{n+1}$$
Therefore for $n>0$, $!n$ is the closest integer to $\frac{n!}{e}$
Source (I didn't derive this): https://math.stackexchange.com/a/83472/623901
Method 3 (inclusion-exclusion):
