# Shifted Factorial

1. Jun 28, 2011

### Ted123

1. The problem statement, all variables and given/known data

I've got to show $$\sum_{n=0}^{\infty} \frac{(a)_n(-1)_n}{(c)_n n!} = \frac{c-a}{c}$$
where
$$\displaystyle (a)_n = \frac{\Gamma(a+n)}{\Gamma(a)} = a(a+1)...(a+n-1)$$
is the shifted factorial (Pochhammer symbol).

3. The attempt at a solution

I've been informed that $$(-1)_n = 0\;\;\;\;\;\;\forall\;\;n\geq 2$$
So the sum has only 2 terms for n=0 and n=1, but what do e.g. $$(-1)_0\,,\,(-1)_1\,,\,(a)_0\,,\,(a)_1$$ equal?

2. Jun 28, 2011

### tiny-tim

Hi Ted123!

look at the definition of (a)n

(a)0 obviously = 1 for all a (including (-1)0 = 1),

and (a)1 = … ?

3. Jun 28, 2011

### Ted123

So would (a)1 = a, and (-1)1 = -1 ?

So the 2 terms of the sum give $$1 - \frac{a}{c} = \frac{c-a}{c}$$
Incidentally would (a)2 = a(a+1), (a)3 = a(a+1)(a+2) etc.?

Last edited: Jun 28, 2011
4. Jun 28, 2011

### tiny-tim

yes yes yes yes and yes