Consider the function $\_2F\_1(5.5, 1, 5;-|x|^2]$ for $x\in \mathbb{R}^n.$ I want to show that this function is positive. I checked that it does not have any roots so can I conclude the inequality by using continuity in $x$ of the function $\_2F\_1(5.5, 1, 5;-|x|^2]$?



How to show the following inequality $_2F_1(5.5, 1, 5;-|x|^2]>0$?
You have the integral representation
$${}\_2F\_1\left(\begin{matrix}11/2,1\\5\end{matrix};-x\right)=\frac 4{x^4}\int\_0^x\frac{(x-t)^3}{(1+t)^{11/2}}\,dt,$$
which follows by expanding $1/(1+t)^{11/2}$ using the binomial theorem and integrating termwise. This should prove the positivity.


Note that this integral is a remainder term in the Taylor expansion of $1/(1+x)^{3/2}$. So your series can be summed explicitly as $1/(1+x)^{3/2}$ minus a Taylor polynomial. This is also easy to see by writing
$$\frac{(11/2)\_k}{(5)\_k}=\frac{4!}{(3/2)\_4}\frac{(3/2)\_{k+4}}{(k+4)!}.$$
Your series is essentially just the binomial series for $1/(1+x)^{3/2}$ minus the first four terms. More explicitly,
$${}\_2F\_1\left(\begin{matrix}11/2,1\\5\end{matrix};-x\right)=\frac{128}{315x^4}\left(\frac 1{(1+x)^{3/2}}-1+\frac 32\,x-\frac{15}8\,x^2+\frac{35}{16}\,x^3\right).$$
Perhaps this fact is also useful to you.


