Shifted Factorial Homework: Showing \sum_{n=0}^{\infty}

[Ted123]

Homework Statement

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).

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?

 

[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 = … ? ​

 

[Ted123]

Hi Ted123!

look at the definition of (a)_n …

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

and (a)₁ = … ? ​

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.?

 

[tiny-tim]

yes yes yes yes and yes