# Writing this series as a hypergeometric series

1. Apr 5, 2012

### Ted123

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

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 $_2 F_1(a,b;c;z)$.

2. Relevant equations

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

3. 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)?

Last edited by a moderator: May 5, 2017
2. Apr 5, 2012

### clamtrox

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

3. Apr 5, 2012

### Ted123

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 $a=b=0$?

4. Apr 5, 2012

### 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?

5. Apr 5, 2012

### Ted123

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