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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.

newtykins
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1 Answers1

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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 $$31035053229546199656252032972759319953190362094566672920420940313$$

Claude Leibovici
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