I am trying to sum the following series, given n is a positive integer, $$ n+n(n-1)+n(n-1)(n-2)+...+n! $$ I think there is a solution, but if not, an approximate result will also be appreciated.
Regards.
$\endgroup$ 101 Answer
$\begingroup$As Open Ball almost says, you have $\displaystyle n! \sum_{k=0}^{n-1} \frac1{k!}$ which is almost $n! \, e = \displaystyle n! \sum_{k=0}^\infty \frac1{k!}$
The difference is $\displaystyle n! \sum_{k=n}^\infty \frac1{k!} = 1+ n! \sum_{k=n+1}^\infty \frac1{k!}$ with the remaining sum being less than $1$
So we can round down and subtract $1$ to give $$\displaystyle n! \sum_{k=0}^{n-1} \frac1{k!} = \lfloor n!\, e\rfloor -1$$
$\endgroup$ 2
