Series - series that looks like the product of hypergeometric functions

Question

I have a series of the form

\begin{equation} \sum_{n=0}^{\infty}\frac{(1)_{n}\,(\alpha_{3}+2)_{n}}{(\alpha_{1}+2)_{n}\, n!}\left(\frac{\beta_{1}}{\beta_{3}}\right)^{n}\, {_{1}}F_{1}(-\alpha_{1}-1-n;\,-\alpha_{3}-1-n;\,\beta_{3}x) \end{equation}

where,

$\alpha_{1}<\alpha_{3}, \quad \{\alpha_{1},\alpha_{3}\}\in \mathbb{Z}^{+} \ \text{or} \ \in \frac{\mathbb{Z}^{+}}{2}$

$\beta_{1}<\beta_{3}, \quad \{\beta_{1},\beta_{3}\}\in \mathbb{R}^{+}$

$0\leq x$

which looks very similar to some of the addition theorems for the confluent hypergeometric function ${_{1}}F_{1}(a,b,x)$. Also, take away the ${_{1}}F_{1}$ function from the series and it can be seen that the sum would the the Gauss hypergeometric function ${_{2}}F_{1}$. Been trying to manipulate this for a while with no clear path forward. Any thoughts are appreciated.

Answer 1

Please check and make (constructive) comments and corrections. I usually get some strange dyslexia while thinking.

We start with a few simple formula:

$\left(\gamma+\psi\right)_{n}=\frac{\Gamma\left(\gamma+\psi+n\right)}{\Gamma\left(\gamma+\psi\right)}$

$\Gamma(z)\cdot\Gamma\left(1-z\right)=\frac{\pi}{sin\left(\pi\cdot z\right)}$

And for future compression/representation:

Lemma 1. $e^{\beta\cdot t}\cdot_{p}F_{q}((a);(b);\lambda\cdot t)={\displaystyle \sum_{k=0}^{\infty}\frac{t^{k}}{k!}\cdot\beta^{k}\cdot _{p+1}F_{q}\left(-k,\left(a\right);\left(b\right);-\frac{\lambda}{\beta}\right)}$

Proof. Consider a general convolution term and evaluate the coefficient of $\frac{t^{k}}{k!}$ $e^{\beta\cdot t}\cdot_{p}F_{q}((a);(b);\lambda\cdot t)={\displaystyle \sum_{k=0}^{\infty}}{\displaystyle \sum_{l=0}^{k}\frac{\left(a\right)_{l}}{\left(b\right)_{l}}\frac{\left(\beta\cdot t\right)^{k-l}}{l!}\cdot\frac{\left(\lambda\cdot t\right)^{l}}{\left(k-l\right)!}}$

${\displaystyle \sum_{k=0}^{\infty}\frac{t^{k}}{k!}}{\displaystyle \sum_{l=0}^{k}}\frac{k!}{l!\cdot\left(k-l\right)!}\cdot\beta^{k}\cdot\frac{\left(a\right)_{l}}{\left(b\right)_{l}}\cdot\left(\frac{\lambda}{\beta}\right)^{l}$

$ {\displaystyle \sum_{k=0}^{\infty}\frac{t^{k}}{k!}}{\displaystyle \sum_{l=0}^{k}}\beta^{k}\cdot\frac{\left(-k,\left(a\right)\right)_{l}}{\left(b\right)_{l}}\cdot\frac{\left(-\frac{\lambda}{\beta}\right)^{l}}{l!}$

QED

Note that the spliting of $\beta$ is soley for the purpose of standard generalized Hypergeometric representation; i.e. don't let $\beta^{k}$ wander off unless you realize the risks.

Now to the derivation

$\Gamma\left(\alpha+2+n\right)=\frac{\pi}{\Gamma\left(1-\left(\alpha+2+n\right)\right)\cdot sin\left(\pi\cdot\left(\alpha+2+n\right)\right)}=\frac{\pi}{\Gamma\left(-\left(\alpha+1+n\right)\right)\cdot sin\left(\pi\cdot\left(\alpha+2+n\right)\right)}$

Thus the front end is:

$\frac{sin\left(\pi\cdot\left(\alpha_{1}+2+n\right)\right)}{sin\left(\pi\cdot\left(\alpha_{3}+2+n\right)\right)}\cdot\frac{\Gamma\left(-\left(\alpha_{1}+1+n\right)\right)}{\Gamma\left(-\left(\alpha_{3}+1+n\right)\right)}$

And the Confluent Hypergeometric Function is $$_{1}F_{1}\left(-\left(\alpha_{1}+1+n\right);-\left(\alpha_{3}+1+n\right);\beta_{3}\cdot x\right)={\displaystyle \sum_{k=0}^{\infty}\frac{\frac{\Gamma\left(-\left(\alpha_{1}+1+n\right)+k\right)}{\Gamma\left(-\left(\alpha_{1}+1+n\right)\right)}}{\frac{\Gamma\left(-\left(\alpha_{3}+1+n\right)+k\right)}{\Gamma\left(-\left(\alpha_{3}+1+n\right)\right)}}\cdot\frac{\left(\beta_{3}\cdot x\right)^{k}}{k!}}$$

Thus we have: $${\displaystyle \sum_{n=0}^{\infty}}{\displaystyle \sum_{k=0}^{\infty}}\frac{sin\left(\pi\cdot\left(\alpha_{1}+2+n\right)\right)}{sin\left(\pi\cdot\left(\alpha_{3}+2+n\right)\right)}\cdot\frac{\Gamma\left(k-\alpha_{1}-1-n\right)}{\Gamma\left(k-\alpha_{3}-1-n\right)}\cdot\left(\frac{\beta_{1}}{\beta_{3}}\right)^{n}\cdot\frac{\left(\beta_{3}\cdot x\right)^{k}}{k!}$$

For integer $\alpha$ the sin()'s basically cancel and the ratio can be replaced with something like $(-1)^{\alpha_{1}-\alpha_{3}}$

A standard representation:

This part is incomplete/incorrect. Please finish or delete it: or I will later. (Hint: you have to add $\frac{(1)}{(p-k)!}$ and some other shaping (I think). Now we start to rewrite it and use the Lemma to get a standard (more or less) convolution representation: Let: $p=k-n;n=k-p$

${\displaystyle \sum_{k=0}^{\infty}}{\displaystyle \sum_{p=0}^{\infty}}\frac{sin\left(\pi\cdot\left(\alpha_{1}+2+n\right)\right)}{sin\left(\pi\cdot\left(\alpha_{3}+2+n\right)\right)}\cdot\frac{\Gamma\left(p-\alpha_{1}-1\right)}{\Gamma\left(p-\alpha_{3}-1\right)}\cdot\left(\frac{\beta_{1}}{\beta_{3}}\right)^{p-k}\cdot\frac{\left(\beta_{3}\cdot x\right)^{k}}{k!}$

${\displaystyle \frac{\Gamma\left(-\alpha_{1}-1\right)}{\Gamma\left(-\alpha_{3}-1\right)}\cdot\left(-1\right)^{\left(\alpha_{1}-\alpha_{3}\right)}\cdot\sum_{k=0}^{\infty}}{\displaystyle \frac{t^{k}}{k!}\sum_{n=0}^{\infty}\beta_{3}^{k}\cdot}\frac{\left(-\alpha_{1}-1\right)_{\left(p\right)}}{\left(-\alpha_{3}-1\right)_{\left(p\right)}}\cdot\left(\frac{\beta_{1}}{\beta_{3}}\right)^{p-k}$