Closed form of $\sum_{k=0}^{\infty} \sum_{m=0}^{\infty} r^m \cdot t^k \binom{m+k}{k} \binom{m+k+1}{k}$ for fixed $r, t$

Question

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.

Answer 1

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).