I wish to sum the following series:
$$ \sum_{r=1}^nr! $$
My initial thought was to first convert this in terms of the gamma function, and sum all of the integrals that this made like so:
$$ \sum_{r=1}^nr! = \sum_{r=1}^n\Gamma(r+1)=\sum_{r=1}^n\int^\infty_0x^re^{-x}\,dx $$
No matter how hard I try, I struggle to find a solution yet Wolfram Alpha managed to find the answer to be $(-1)^{n+1}\Gamma(n+2)!(-n-2)-!(-1)-1$ where $!n$ is the sub-factorial function. I feel like this is a pretty complex problem, but could someone try and help me understand this? Or at least give me a better angle to take this problem from as using the gamma function may not be the best way to solve this.
Beside the results given in the linked answers, we can write $$\sum_{r=1}^n r!\sim n! \sum_{k=0}^\infty \frac {B_k }{n^k}$$ where $B_k$ is a Bell number.
You can approximate the infinite summation by its $[m,m]$ Padé approximant $P_m$, the first ones being $$P_1=\frac{n}{n-1}\qquad \qquad P_2=\frac{n^2-2 n-1}{n^2-3 n+1}\qquad \qquad P_3=\frac{n^3-5 n^2+3 n+3}{n^3-6 n^2+8 n-1}$$ $$P_4=\frac{n^4-9 n^3+20 n^2-3 n-9}{n^4-10 n^3+29 n^2-24 n+1}\qquad \qquad P_5=\frac{n^5-14 n^4+61 n^3-83 n^2-6 n+33}{n^5-15 n^4+75 n^3-145 n^2+89 n-1}$$
Using $P_5$ for $n=50$ will give as an approcimation $$\sum_{r=1}^{50} r!\sim \frac{232417233}{227766949}\, (50)!$$ which, rounded, is $$\color{red}{3103505322954619}8048161042513108335457375157009824023613291215487$$ to be compared to the exact value $$31035053229546199656252032972759319953190362094566672920420940313$$