# Is this a new formula?

It's late at night and I'm tired, but I just stumbled across this while doing my homework. Any chance this is new? Or, maybe, did I just somehow transform it and it is still basically the same formula? In that case, forgive me, please.

Anyway, here it is. It is a recursive function (thus, of limited use!?) to calculate the sum of factorials:

$f(n) = \sum\limits_{i=1}^n i! = \frac{(n+1)!}{n}+f(n-2)$

with

$f(0) = 0$ and $f(-1) = -1$

Is this useful at all or did I just waste my time? :)

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The formula certainly is true. Re-arranging what you wrote you get

$$f(n) - f(n-2) = n! + (n-1)! = n(n-1)! + (n-1)!$$

$$= (n+1)\times (n-1)! = \frac{(n+1)\times n\times (n-1)!}{n} = \frac{(n+1)!}{n}$$

but I don't think it is particularly more useful than the observation of, say,

$$f(n) = n! + f(n-1)$$

The definition of your function as a sum means that it naturally can be described recursively. I don't think the formulation you gave offers any particular advantage to the summation formula, unfortunately.

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Ah, too bad. Thanks for the writeup. –  Franz Jan 25 '11 at 0:11