Writing this series as a hypergeometric series

[Ted123]

Homework Statement

Write \displaystyle \sum_{k=0}^{\infty} \frac{1}{9^k (\frac{2}{3})_k} \frac{w^{3k}}{k!} in terms of the Gauss hypergeometric series of the form _2 F_1(a,b;c;z).

Homework Equations

The Gauss hypergeometric series is http://img200.imageshack.us/img200/5992/gauss.png

The Attempt at a Solution

It's nearly a series of that form if I put z=w^3 and k=n but how do I get the 9^{-k} = 3^{-k}3^{-k} factors in terms of shifted factorials (that is if I need to)?

 

[clamtrox]

That term does not go into the factorials, it goes into z^n.

 

[Ted123]

That term does not go into the factorials, it goes into z^n.

Ah of course. So if I put z=\frac{w^3}{9} then the series can be written as _2 F_1 (a,b ; \frac{2}{3} ; \frac{w^3}{9}) for some a and b with (a)_n(b)_n = 1 for all n=0,1,2,... Can I just pick a=b=0?

 

[clamtrox]

I would do some extra checking to be sure that that's right. Can you plug the solution into the hypergeometric differential equation with a=b=0 and see if it solves it?

 

[Ted123]

I would do some extra checking to be sure that that's right. Can you plug the solution into the hypergeometric differential equation with a=b=0 and see if it solves it?

Actually (0)_n (0)_n \neq 1 for n=0,1,2,... so how do I get 2 shifted factorials to equal 1?