To obtain a closed form for the series related to special functions.

Question

I am learning the properties of special functions particularly, hypergeometric functions. I got the following series form: $$f(z)=\Gamma(3)\left[\frac{z}{\Gamma(3)}+\frac{(\gamma)_1}{\Gamma(4)}\frac{z^2}{1!}+\frac{(\gamma)_{2}}{\Gamma(5)}\frac{z^3}{2!}+\frac{(\gamma)_{3}}{\Gamma(6)}\frac{z^4}{3!}+\cdots\right].$$ I would like to obtain a closed form for the above infinite series. Here is what I have done. On the right hand side, multiplied and divided by $z$ to get $$f(z)=\frac{\Gamma(3)}{z}\left[\frac{z^2}{\Gamma(3)}+\frac{(\gamma)_1}{\Gamma(4)}\frac{z^3}{1!}+\frac{(\gamma)_{2}}{\Gamma(5)}\frac{z^4}{2!}+\frac{(\gamma)_{3}}{\Gamma(6)}\frac{z^5}{3!}+\cdots\right],$$ which can be further written as $$f(z)=\frac{2}{z}\left[\frac{z^2}{2!}+\frac{(\gamma)_1}{3!}\frac{z^3}{1!}+\frac{(\gamma)_{2}}{4!}\frac{z^4}{2!}+\frac{(\gamma)_{3}}{5!}\frac{z^5}{3!}+\cdots\right].$$ Using $(1)_n=n!$, I got $$f(z)=\frac{2}{z}\left[\frac{z^2}{2!}+\frac{(\gamma)_1}{(1)_1}\frac{z^3}{3!}+\frac{(\gamma)_{2}}{(1)_2}\frac{z^4}{4!}+\frac{(\gamma)_{3}}{(1)_3}\frac{z^5}{5!}+\cdots\right].$$ This seems to be related to confluent hypergeometric function. I don't know to how to proceed further. Could someone provide me hints? Thanks.

Answer 1

You are right, the series is a little messed up, but there's no reason to try guessing the proper form.

We can always use the general way to get the hypergeometric form of a series. Consider:

$$f(z)=z \Gamma(3)\left[\frac{1}{\Gamma(3)}+\frac{(\gamma)_1}{\Gamma(4)}\frac{z}{1!}+\frac{(\gamma)_{2}}{\Gamma(5)}\frac{z^2}{2!}+\frac{(\gamma)_{3}}{\Gamma(6)}\frac{z^3}{3!}+\cdots\right]=2 z \sum_{n=0}^\infty \frac{(\gamma)_n z^n}{(n+2)! n!}$$

Now we catch the $0$ term:

$$A_0= \frac{2z}{2}=z$$

And consider the ratio of terms:

$$\frac{A_{n+1}}{A_n}= \frac{(n+\gamma)}{(n+3)} \frac{z}{n+1}$$

Which immediately allows us to write:

$$f(z)=z {_1 F_1} (\gamma;3;z)$$