I'm trying to prove the sum of a sequence given by


$a\_{n+1} = \frac{nb-x}{(n+1)b} a\_n$


with $a\_1 = 1$. This gives the solution $a\_n = \frac{(-x/b)\_n}{n!}$. When trying to work out what this sums to, I looked at hypergeometric functions ${}\_1F\_0(-x/b;;1)$ to sum this, but this appears to be undefined.
I have another results saying that $a\_1 = x/b$, and Mathematica seems to agree, 


Sum[Pochhammer[1/3, n - 1]/Gamma[1 + n], {n, 1, Infinity}] = 3/2
but I can't see why.


Any help??



Sum of difference equation involving hypergeometric functions 1F0
The solution you give for $a\_{n+1} = \frac{nb-x}{(n+1)b} a\_n$ has a missing factor, I think it should read
$$a\_n=-\frac{b}{x}\frac{(-x/b)\_n}{n!}.$$ 
Then the sum over $n$ equals 
$$\sum\_{n=1}^\infty a\_n =\frac{b}{x},$$
as follows from the generating function
$$\sum\_{n=0}^\infty (p)\_n\frac{z^n}{n!}=(1-z)^{-p}$$
(here is a [derivation](http://www.mi.sanu.ac.rs/~gvm/radovi/AP-JIS2.pdf) --- page 2)


