# Can we confirm $\sum_x{x!}$?

According to Wikipedia's article on indefinite sums, they list the following formula near the bottom of the page:

$$\displaystyle \sum_x{\Gamma(x)}=(-1)^{x+1}\Gamma(x)\frac{\Gamma(1-x,-1)}{e}+C$$

However, in Mathematica 7.0.1, I get the following:

$$\displaystyle\sum_x{(x-1)!} =$$ $$\displaystyle \sum_x{\Gamma(x)} = (-1)^{x+2}\Gamma(x+1)\frac{\Gamma(1-(x+1),-1)}{e}+ (-1)^{2}\Gamma(1)\frac{\Gamma(0,-1)}{e}$$

Where I've substituted $(-1)^{2}\Gamma(1)\frac{\Gamma(0,-1)}{e}$ for $C$.

Can someone please confirm that my equations are correct? I'm using this formula for an important algorithm, and I'd like to be certain that the math checks out. Thanks!

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The wiki page is correct, its difference delta is $\Gamma(x)$. The other answer's difference delta is $\Gamma(x+1) = x!$. –  Sasha Nov 2 '11 at 14:35
@Sasha: Thanks for correcting my math. –  Matt Groff Nov 2 '11 at 14:43
Well in Mathematica I got $$\sum_{x=0}^{n}x!=-(-1)^n \Gamma (n+2) \text{Subfactorial}[-n-2]-\text{Subfactorial}[-1]-1$$ –  GarouDan Nov 4 '11 at 22:10