$\sum_{i=1}^n P(n,i)$

Question

We can know clearly from

$$(1+X)^n=\sum\limits_{i=0}^{n}C(n,i)X^n$$

that

$$ \sum\limits_{i=0}^{n}C(n,i)=2^n.$$

Whereas, I want to know if there are any researched results about permutations in the similar case, i.e., what can we know about

$$ \sum\limits_{i=0}^{n}P(n,i).$$

I’m really curious about that, but have found no answers elsewhere. Any help will be sincerely appreciated!

Answer 1

$\sum\limits_{i=0}^{n}P(n,i) = \sum\limits_{i=0}^{n}\frac{n!}{(n-i)!}= \sum\limits_{i=0}^{n}\frac{n!}{i!}=n!\sum\limits_{i=0}^{n}\frac{1}{i!}=n!e - n!\sum\limits_{i=n+1}^{\infty}\frac{1}{i!} $

so

$n!e - \sum\limits_{i=0}^{n}P(n,i) = n!\sum\limits_{i=n+1}^{\infty}\frac{1}{i!} < \sum\limits_{i=1}^{\infty}\frac{1}{(n+1)^i}$

so

$n!e - \sum\limits_{i=0}^{n}P(n,i) < \frac{1}{n}$

Since $\sum\limits_{i=0}^{n}P(n,i)$ is an integer, we have

$\sum\limits_{i=0}^{n}P(n,i) = \lfloor {n!e} \rfloor $