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}


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

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

$0\leq x$

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.

  • $\begingroup$ Well, there is a simplification by doing a "Metamorphosis" on the front end. For the sake of simplicity: are the a,d real or integer. Integers are substantially easier. $\endgroup$
    – rrogers
    Aug 3, 2016 at 14:17
  • $\begingroup$ @rrogers I updated the question with the exact series and clarified the assumptions. All parameters are real. $\alpha_{1}$ and $\alpha_{3}$ are either both positive integers, both positive half integers, or some combination of the two. I know the integer case reduces the 1F1 to a polynomial which should as you said make it a simpler sum. I am certainly interested in the special case of $\alpha_{1}$ and $\alpha_{3}$ being both integers. That said, if $\alpha_{1}$ and $\alpha_{3}\in \frac{\mathbb{Z}^{+}}{2}$, the kummer transformation can make 1F1 into poly with exp term as a constant. $\endgroup$ Aug 3, 2016 at 15:00
  • $\begingroup$ @rrogers I also happen to have a closed-form solution for the derivative of the above function w.r.t. $x$. If that helps. $\endgroup$ Aug 3, 2016 at 15:12
  • $\begingroup$ I would like to see the derivative. I finally got around to running tests in reduce-algebra ( a symbolic math program) and my result seems correct; although I didn't carry it all the way through. reduce-algebra also coughed up a closed form of hypergeometric function in terms of pochammer(x,n/2) (although I can usually change the n/2 results to n). Be aware you can't completely trust these answers and the answer, for general expressions, comes with caveats. $\endgroup$
    – rrogers
    Sep 15, 2016 at 11:07

1 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:


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

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!}$


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:


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: $p=k-n;n=k-p$

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

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

  • $\begingroup$ If you could please show how you got to this point that would be great. I have a few series similar to this one that probably can be simplified the same way you did here. Thank you for your time. $\endgroup$ Aug 3, 2016 at 22:03
  • $\begingroup$ Thank you for your answer. The explanation of your derivation really helps. As for the last bit of your post, I would leave it as is (or maybe bold the bit about it being incomplete or incorrect). It's still a potentially useful approach for moving forward with a final solution. Thank you again for your time. $\endgroup$ Aug 4, 2016 at 15:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.