I'm struggling with the following summation $$F(x)=\sum\limits_{t=1}^{\infty}\binom{-5/2}{2t}\binom{2t}{t-1}x^{t}$$ I have tried to look for ways to reduce the product of the binomial coefficient to no avail.
Any hints or suggestions would be much appreciated.
If Wolfram Alpha is anything to go by, the best we can likely do is write it in terms of the hypergeometric function, which can at least be efficiently computed. We'll use Wikipedia's notation from their treatement of the hypergeometric series, with $(q)_n$ the rising Pochhammer symbol (although unfortunately, in their article about the symbol this instead denotes the falling one). Write $t=k+1$ so $F=\frac{35x}{8}G$ with$$G:=\sum_{k\ge0}\frac{\frac{8}{35}\left(-\frac52\right)!}{k!(k+2)!\left(-\frac92-2k\right)!}x^k=\sum_k\frac{\left(\frac94\right)_k\left(\frac{11}{4}\right)_k}{(3)_k}\frac{(4x)^k}{k!}={}_2F_1\left(\frac94,\,\frac{11}{4};\,3;\,4x\right).$$In particular, $(k+2)!=2(3)_k$ and$$\begin{align}\frac{\frac{4}{35}\left(-\frac52\right)!}{\left(-\frac92-2k\right)!}&=\frac{\left(-\frac92\right)!}{\left(-\frac92-2k\right)!}\\&=\prod_{j=0}^{2k-1}(\tfrac92+j)\\&=4^k\prod_{j=0}^{2k-1}(\tfrac94+\tfrac{j}{2})\\&=4^k\prod_{j=0}^{k-1}(\tfrac94+j)\prod_{j=0}^{k-1}(\tfrac{11}{4}+j)\\&=4^k(\tfrac94)_k(\tfrac{11}{4})_k.\end{align}$$