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I would like to find the closed form for the double sum $$\sum_{k=0}^{\infty} \sum_{m=0}^{\infty} r^m \cdot t^k \binom{m+k}{k} \binom{m+k+1}{k} \tag 1$$

where $r, t$ are known values. When I plugged this into Mathematica, I got two equivalent sums: $$\sum_{m=0}^{\infty} r^m \space _2F_1(1+m, 2+m, 1, t) = \tag 2$$ $$\sum_{k=0}^{\infty} (1+k)t^k \space _2F_1(1+k, 2+k, 2, r) \tag 3$$

Here, $_2F_1$ is a hypergeometric function. What I noticed is that $_2F_1(1+m, 2+m, 1, t) = \frac{P_m}{(1-t)^{2m+2}}$, where $P_m$ is a polynomial in $t$ of degree $m$ and that $_2F_1(1+k, 2+k, 2, r) = \frac{Q_k}{(1-r)^{2k+1}}$, where $Q_k$ is a polynomial in $r$ of degree $k-1$.

I tried taking another approach, changing the indices of $(1)$ so that $s = k+m$: $$\sum_{s=0}^{\infty} \sum_{k=0}^{s} r^{s-k} \cdot t^k \binom{s}{k} \binom{s+1}{k} \tag 4$$

This however, also led to a sum of hypergeometric functions: $$\sum_{s=0}^{\infty} r^s \space _2F_1 \left( -1-s, -s, 1, \frac{t}{r} \right) \tag 5$$

All these attempts seem pointless to me, since using the hypergeometric function just yields a more compact way of expressing the double sum (not an actual simplification), which brings me to the main question: How can I get a closed form of the original double sum?

Edit: $r, t < 0$ if it makes any difference.

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    $\begingroup$ Are we sure there's a closed form at all? $\endgroup$ Apr 15, 2020 at 22:53
  • $\begingroup$ There is no reason to be sure that there is a closed form, but I'm hoping there is. $\endgroup$ Apr 16, 2020 at 3:22

2 Answers 2

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An expression of this double sum in terms of a special function can be obtained as \begin{align} f(t,r)&=\sum_{k=0}^{\infty} \sum_{m=0}^{\infty} r^m t^k \binom{m+k}{k} \binom{m+k+1}{k} \\ &=\sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \frac{(1)_{m+k}(2)_{m+k}}{(1)_{k}(2)_{m}}\frac{ t^kr^m}{k!m!} \end{align} Using the definition of the fourth Appell function \begin{equation} {F_{4}}\left(\alpha,\beta;\gamma,\gamma^{\prime};x,y\right)=\sum_{m,n=0}^{% \infty}\frac{{\left(\alpha\right)_{m+n}}{\left(\beta\right)_{m+n}}}{{\left(% \gamma\right)_{m}}{\left(\gamma^{\prime}\right)_{n}}m!n!}x^{m}y^{n} \end{equation} valid for $\sqrt{\left|x\right|}+\sqrt{\left|y\right|}<1$, we identify \begin{equation} f(t,r)={F_{4}}\left(1,2;1,2;t,r\right) \end{equation} when $\sqrt{\left|r\right|}+\sqrt{\left|t\right|}<1$. This function cannot be expressed as the product of two hypergeometric functions, in general. A list of its properties (including the OP expressions (2) and (3)) can be found in an article by Brychkov and Saad (under a paywall).

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As usual; check for typos before using. These were checked on a maxima worksheet. Which I will clean up and supply if requested.
The "power tools" are described in some detail since I haven't seen them on this forum. They have a lot of derivatives floating around the internet; in various forms and for various purposes.
There are some questions one might regret asking and answering :)
The answer is:
$$G\left(u\right)$$ $$=-\frac{\left(t-r\right)\sqrt{\left({{t}^{2}}-2rt+{{r}^{2}}\right){{u}^{2}}+\left(-2t-2r\right)u+1}+\left({{t}^{2}}-2rt+{{r}^{2}}\right)u-t-r}{\left(\left(rt-{{r}^{2}}\right)u+r\right)\sqrt{\left({{t}^{2}}-2rt+{{r}^{2}}\right){{u}^{2}}+\left(-2t-2r\right)u+1}+\left(r{{t}^{2}}-2{{r}^{2}}t+{{r}^{3}}\right){{u}^{2}}+\left(-2rt-2{{r}^{2}}\right)u+r}$$
Where the actual summation would be done by setting $u=1$ or the individual terms by $\left[u^{n}\right]G\left(u\right)$. "If you see me, then weep"

https://en.wikipedia.org/wiki/Hunger_stone
We use some “power tools”. There are two of them:

(I have changed variable names to simplify reading)

  1. Euler's Combinatorial Transform; from https://web.archive.org/web/20120127210623/http://www.dsi.unifi.it/~resp/GouldBK.pdf

Equations (1) and (2)

(1) Let: $$\phi\left(u\right)={\displaystyle \sum_{k=0}^{\infty}\phi_{k}\cdot u^{k}}$$
(2) Then:$$\left[u^{s}\right]\frac{1}{1-r\cdot u} \phi\left(\frac{1}{1-r\cdot u}\right)={\displaystyle \left({\displaystyle \sum_{k=0}^{s}r^{s-k}\left(\begin{array}{c} s\\ k \end{array}\right)\phi_{k}}\right)}$$
In our case we have to slightly rewrite; absorb the problem $t$ into $\phi\left(u\right)$ with $u$ being the indexing variable
(1): $$\phi\left(u\right)={\displaystyle \sum_{k=0}^{\infty}\phi_{k}\cdot t^{k}\cdot u^{k}}$$
(2): $$\left[u^{s}\right]\frac{1}{1-r\cdot u}\phi\left(\frac{1}{1-r\cdot u}\right)={\displaystyle \left({\displaystyle \sum_{k=0}^{s}r^{s-k}\left(\begin{array}{c} s\\ k \end{array}\right)\phi_{k}\cdot t^{k}}\right)}$$ 2) The Diagonlization rule, if applicable (our case):

https://www.sciencedirect.com/science/article/pii/S0012365X07010345

Theorem 2: Let: $$\phi(0)\ne0,\phi\left(u\right)={\displaystyle \sum_{k=0}^{\infty}\phi_{k}\cdot t^{k}\cdot u^{k}}$$
$$g_{s}=\left[u^{s}\right]f(u)\phi(u)^{s}$$
$$i.e.\,G(u)=\sum_{k=0}^{\infty}g_{k}\cdot u^{k}$$
$$w\left(u\right)=u\cdot\phi\left(w\right)$$
Then, if the last equation is solvable:
$$G\left(u\right)=\left[\frac{f\left(w\left(u\right)\right)}{1-u\cdot\phi'(w\left(u\right))}\right]$$
Where $$\phi'(t,w)=\frac{\partial\phi(t,w)}{\partial w}$$
—————————-
Our problem: Eq: 4
$${\displaystyle \sum_{s=0}^{\infty}}\sum_{k=0}^{s}r^{s-k}\cdot\left(\begin{array}{c} s\\ k \end{array}\right)\cdot\left(\begin{array}{c} s+1\\ k \end{array}\right)t^{k}$$
$$\phi(u)=\left(1+u\cdot t\right)^{s+1}=\sum_{k=0}^{s+1}\left(\left(\begin{array}{c} s+1\\ k \end{array}\right)t^{k}\right)u^{k}=\phi_{k}u^{k}$$
Applying Euler's Transform
$${\displaystyle \sum_{k=0}^{s}\left(\begin{array}{c} s\\ k \end{array}\right)r^{n-k}\phi_{k}=\sum_{k=0}^{s}\left(\begin{array}{c} s\\ k \end{array}\right)r^{n-k}\left(\begin{array}{c} s+1\\ k \end{array}\right)t^{k}=\left[u^{s}\right]\frac{1}{1-u\cdot r}\phi\left(t\cdot\frac{u}{1-u\cdot r}\right)}$$
We have
$$\left[u^{s}\right]\frac{1}{1-u\cdot r}\left(1+t\cdot\frac{u}{1-u\cdot r}\right)^{s+1}$$
$$=\left[u^{s}\right]\frac{\left(1+u\left(t-r\right)\right)}{\left(1-r\cdot u\right)^{2}}\left(\frac{1+u\cdot\left(t-r\right)}{\left(1-r\cdot u\right)}\right)^{s}$$
Which is diagonal
$$f\left(u\right)=\frac{\left(1+u\left(t-r\right)\right)}{\left(1-r\cdot u\right)^{2}};\,\,\phi(u)=\left(\frac{1+u\cdot\left(t-r\right)}{\left(1-r\cdot u\right)}\right)$$
$$solve\left(w\left(u\right)=u\cdot\left(\frac{1+w\left(u\right)\cdot\left(t-r\right)}{\left(1-r\cdot w\left(u\right)\right)}\right), w\right)$$
$$w=\pm\frac{\sqrt{\left(t^{2}-2\cdot r\cdot t+r^{2}\right)\cdot u^{2}-2\cdot u\cdot\left(t-r\right)+1}\pm\left(\left(r-t\right)\cdot u+1\right)}{2r}$$
We use
$$w=-\frac{\sqrt{\left(t^{2}-2\cdot r\cdot t+r^{2}\right)\cdot u^{2}-2\cdot u\cdot\left(t-r\right)+1}-\left(\left(r-t\right)\cdot u+1\right)}{2r}$$
Evaluting the diagonalization terms
$$f\left(w\left(u\right)\right)=\frac{\left(1+w\left(u\right) \left(t-r\right)\right)}{\left(1-r\cdot w\left(u\right)\right)^{2}}$$
$$\phi(w\left(u\right))=\left(\frac{1+w\left(u\right)\cdot\left(t-r\right)}{\left(1-r\cdot w\left(u\right)\right)}\right)$$
$$\phi'(w\left(u\right))=\frac{r\cdot\left(t\cdot w\left(u\right)-r\cdot w\left(u\right)+1\right)}{\left(r\cdot w\left(u\right)-1\right)^{2}}-\frac{t-r}{r\cdot w\left(u\right)-1}$$
And we express
$$G\left(u\right)=\left[\frac{f\left(w\left(u\right)\right)}{1-u\cdot\phi'(w\left(u\right))}\right]$$
$$G\left(u\right)=\frac{\left(t-r\right)\cdot w\left(u\right)+1}{\left(r\cdot w\left(u\right)\right){}^{2}-2r\cdot w\left(u\right)-t\cdot u+1}$$
$$G\left(u\right)=-\frac{\left(t-r\right)\sqrt{\left({{t}^{2}}-2rt+{{r}^{2}}\right){{u}^{2}}+\left(-2t-2r\right)u+1}+\left({{t}^{2}}-2rt+{{r}^{2}}\right)u-t-r}{\left(\left(rt-{{r}^{2}}\right)u+r\right)\sqrt{\left({{t}^{2}}-2rt+{{r}^{2}}\right){{u}^{2}}+\left(-2t-2r\right)u+1}+\left(r{{t}^{2}}-2{{r}^{2}}t+{{r}^{3}}\right){{u}^{2}}+\left(-2rt-2{{r}^{2}}\right)u+r}$$
Here are some results from Maxima
$$\left[u^{5}\right]G\left(u\right)=6t^{5}+75rt^{4}+200r^{2}t^{3}+150r^{3}t^{2}+30r^{4}t+r^{5}$$
Done with sum command
$$\sum_{k=0}^{5}\left(\begin{array}{c} 5\\ k \end{array}\right)r^{n-k}\left(\begin{array}{c} 5+1\\ k \end{array}\right)t^{k}=6t^{5}+75rt^{4}+200r^{2}t^{3}+150r^{3}t^{2}+30r^{4}t+r^{5}$$
And Euler result

$$\left[u^{5}\right]\frac{1}{1-u\cdot r}\left(1+t\cdot\frac{u}{1-u\cdot r}\right)^{5+1}=6t^{5}+75rt^{4}+200r^{2}t^{3}+150r^{3}t^{2}+30r^{4}t+r^{5}$$

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    $\begingroup$ Interesting derivation. (+1) I'll have a closer look at it soon. This technique is also known as Lagrange inversion. An example can be found e.g. here. $\endgroup$ Oct 11, 2022 at 19:21
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    $\begingroup$ Thanks for the pointers. I had looked at that paper some time ago and thought it was too specialized. It seems I was wrong. I am thinking about editing this. Upon reflection, the binomial( s+1,k) could have been binomial(s+r,k), and using Maxima's "simplify" command at the last was an error; just writing out the expression without "simplification" would have been better. I will reread local.disia.unifi.it/merlini/papers/LagrangeInversion.pdf In light of actually using it. $\endgroup$
    – rrogers
    Oct 11, 2022 at 20:52

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