Gauss hypergeometric series

[Ted123]

Homework Statement

Express $$\sum_{n=0}^{\infty} \frac{1}{(\frac{2}{3})_n} \frac{(z^3/9)^n}{n!}$$ in terms of the Gauss hypergeometric series.

Homework Equations

The Gauss hypergeometric series has 3 parameters a,b,c: $$_2 F_1 (a,b;c;z) = \sum_{n=0}^{\infty} \frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!}$$

The Attempt at a Solution

I can express it as a hypergeometric series: $_0 F_1 (-,-;\frac{2}{3};\frac{z^3}{9})$ but my question is specific in saying it wants it expressed in terms of the Gauss hypergeometric series which has 3 parameters, not 1. I'm stuck as how to do this?

 

[mighty2000]

Hello,

Thank you for your question. Expressing the given series in terms of the Gauss hypergeometric series can be done by rewriting the given series using the properties of the Pochhammer symbol (also known as the rising factorial).

First, let's rewrite the given series:

\sum_{n=0}^{\infty} \frac{1}{(\frac{2}{3})_n} \frac{(z^3/9)^n}{n!} = \sum_{n=0}^{\infty} \frac{1}{\frac{2}{3}\cdot \frac{5}{3}\cdot \frac{8}{3} \cdots \frac{3n+2}{3}} \frac{(z^3/9)^n}{n!}

Now, using the property (a)_n = \frac{\Gamma(a+n)}{\Gamma(a)} and the fact that \Gamma(a+1) = a\Gamma(a), we can simplify the Pochhammer symbol in the denominator as follows:

\frac{1}{\frac{2}{3}\cdot \frac{5}{3}\cdot \frac{8}{3} \cdots \frac{3n+2}{3}} = \frac{\Gamma(\frac{3}{2}+n)}{\Gamma(\frac{3}{2})} \cdot \frac{\Gamma(\frac{5}{2}+n)}{\Gamma(\frac{5}{2})} \cdot \frac{\Gamma(\frac{8}{2}+n)}{\Gamma(\frac{8}{2})} \cdots \frac{\Gamma(\frac{3n+2}{2}+n)}{\Gamma(\frac{3n+2}{2})}

Now, using the property \Gamma(a+1) = a\Gamma(a), we can write the above expression as:

\frac{1}{\frac{2}{3}\cdot \frac{5}{3}\cdot \frac{8}{3} \cdots \frac{3n+2}{3}} = \frac{\frac{3}{2}+n}{\frac{3}{2}} \cdot \frac{\frac{5}{2}+n}{\frac{5}{2}} \cdot \frac{\frac{8}{2}+n}{\frac{8}{2}} \cdots \