I don't have experience with hypergeoemtric functions, but wish to show $$_1F_2[2; \frac{9}{4}, \frac{11}{4}; -\frac{x^2}{16}] > 0 , \text{ when } x\in(0,5).$$ I used Maple to plot the graph and it does greater than $0$ when $x\in(0,5)$. For the proof, my idea is to use asymptotic expansion to show it but I didn't get the result I wanted.
Any help would be appreciated. Thanks!!
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This is not a proof since based on numerical calculations.
If we consider the function $$f(x)=\, _1F_2\left(2;\frac{9}{4},\frac{11}{4};-\frac{x^2}{16}\right)$$ the plot reveals that it is positive up to the first root which is $\approx 10.6155$.
If we expand its as truncated series up to $O(x^{2n})$ and solve for the first positive root $x_{2n}$ using Newton method, we get $$\left( \begin{array}{cc} n & x_{2n} \\ 1 & 7.03562 \\ 3 & 9.30115 \\ 5 & 10.4850 \\ 6 & 10.6338 \\ 7 & 10.6139 \\ 8 & 10.6156 \\ 9 & 10.6155 \end{array} \right)$$ For $n=2$ and $n=4$, there is no real solution to $f(x)=0$.
What we can also do is to build Padé approximants around $x=0$. A simple one, which matches quite well the function $f(x)$ is $$g(x)=\frac{1-\frac{102272 }{7156611}x^2+\frac{139136 }{2799746235} x^4}{1+\frac{1282 }{216867}x^2+\frac{961 }{70048041}x^4 }$$ which shows a first root at $x\approx 10.9691$.
Since the argument $-x^2/16$ is negative, the expansion of the hypergeometric function at zero is an alternating series, with the ratio between the successive terms given by $$\frac {a_{k+1}} {a_k} = -\frac {k+2} {(k+1)(4k+9)(4k+11)} x^2.$$ For a fixed $x \in \left[0, \sqrt {99/2} \,\right)$, the absolute values of $a_k$ go to zero monotonically, and the error can be estimated by the first discarded term: $$\left| {_1F_2} \left(2; \frac 9 4, \frac {11} 4; -\frac {x^2} {16} \right) - 1 \right| \leq \frac {2x^2} {99},$$ which implies that the function is positive on $\left[0, \sqrt {99/2} \,\right)$.