Sum of difference equation involving hypergeometric functions 1F0

Question

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??

Answer 1

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 --- page 2)